[THOUGHT] We need to analyze the surface S given by parametrization: [ \begin{pmatrix}x\y\z\end{pmatrix} = R(\theta) \begin{pmatrix}\cos\phi + 2 \0 \ \sin\phi\end{pmatrix} ] where ( R(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta & 0 \ \sin\theta & \cos\theta & 0 \ 0 & 0 & 1 \end{pmatrix} ) is a rotation about the z-axis by angle θ.

So: [ x = (\cos\phi + 2)\cos\theta ] [ y = (\cos\phi + 2)\sin\theta ] [ z = \sin\phi ]

This is a torus: the circle ( (\cos\phi+2, 0, \sin\phi) ) in the xz-plane (radius 1, center (2,0,0)) rotated about z-axis.

Region V: interior of the torus. Region W: solid cylinder ( x^2 + y^2 \leq 4 ).

We need:

I. Inward unit normal vector at P = ( (1/\sqrt{2}, 1/\sqrt{2}, 0) ) on S.

First, find θ, φ for P.

We have: ( x = (\cos\phi+2)\cos\theta = 1/\sqrt{2} ) ( y = (\cos\phi+2)\sin\theta = 1/\sqrt{2} ) ( z = \sin\phi = 0 )

From z=0 → φ=0 or φ=π. But φ in [0,2π).

Check φ=0: cosϕ=1 → cosϕ+2=3. Then x=3 cosθ = 1/√2, y=3 sinθ = 1/√2 → cosθ=sinθ=1/(3√2). But cosθ=sinθ implies θ=π/4+2kπ? Actually cosθ=sinθ → tanθ=1 → θ=π/4, 5π/4. Check cosθ=1/√2? Wait cosθ=1/(3√2) not 1/√2. So φ=0 gives cosθ=1/(3√2) not matching 1/√2 unless 3=1? No.

Try φ=π: cosϕ=-1 → cosϕ+2=1. Then x=1·cosθ=1/√2, y=1·sinθ=1/√2 → cosθ=sinθ=1/√2 → θ=π/4.

Thus φ=π, θ=π/4.

So P corresponds to (θ=π/4, φ=π).

Now surface S: F(θ,φ) = ((cosφ+2)cosθ, (cosφ+2)sinθ, sinφ).

Compute partial derivatives:

( F_θ = \frac{\partial F}{\partial θ} = (-(cosφ+2)sinθ, (cosφ+2)cosθ, 0) )

( F_φ = \frac{\partial F}{\partial φ} = (-\sinφ \cosθ, -\sinφ \sinθ, \cosφ) )

Normal vector N = F_θ × F_φ.

Compute cross product:

Let A = F_θ = ( -a sinθ, a cosθ, 0 ) where a = cosφ+2. Let B = F_φ = ( -sinφ cosθ, -sinφ sinθ, cosφ ).

Cross product N = A × B:

i-component: (a cosθ)(cosφ) - (0)(-sinφ sinθ) = a cosθ cosφ.

j-component: (0)(-sinφ cosθ) - (-a sinθ)(cosφ) = a sinθ cosφ? Wait compute carefully:

N = | i j k | | -a sinθ a cosθ 0 | | -sinφ cosθ -sinφ sinθ cosφ |

i: (a cosθ)(cosφ) - (0)(-sinφ sinθ) = a cosθ cosφ.

j: (0)(-sinφ cosθ) - (-a sinθ)(cosφ) = a sinθ cosφ.

But careful: j-component formula: - ( (-a sinθ)(cosφ) - (0)(-sinφ cosθ) )? Actually cross product formula: N = (A2 B3 - A3 B2, A3 B1 - A1 B3, A1 B2 - A2 B1).

Let’s compute systematically:

A1 = -a sinθ, A2 = a cosθ, A3 = 0. B1 = -sinφ cosθ, B2 = -sinφ sinθ, B3 = cosφ.

N1 = A2 B3 - A3 B2 = (a cosθ)(cosφ) - (0)(-sinφ sinθ) = a cosθ cosφ. N2 = A3 B1 - A1 B3 = (0)(-sinφ cosθ) - (-a sinθ)(cosφ) = a sinθ cosφ. N3 = A1 B2 - A2 B1 = (-a sinθ)(-sinφ sinθ) - (a cosθ)(-sinφ cosθ) = a sinθ sinφ sinθ + a cosθ sinφ cosθ = a sinφ (sin²θ + cos²θ) = a sinφ.

Thus N = (a cosθ cosφ, a sinθ cosφ, a sinφ).

At θ=π/4, φ=π: a = cosπ+2 = -1+2=1. cosφ = -1, sinφ=0.

So N = (1·cos(π/4)·(-1), 1·sin(π/4)·(-1), 1·0) = (-1/√2, -1/√2, 0).

Magnitude: ||N|| = sqrt( (1/2)+(1/2)+0 ) = 1.

Thus unit normal n = N/||N|| = (-1/√2, -1/√2, 0).

We need inward normal: interior V is inside torus. For a torus, inward normal points toward interior of torus. At P, is N inward? Let’s check: P is on surface, interior is inside torus (the region V). The torus is like a doughnut; interior is the solid doughnut. At P, coordinates (1/√2,1/√2,0) ≈ (0.707,0.707,0). The torus center is at (2,0,0)? Actually the torus is generated by circle centered at (2,0,0) radius 1 in xz-plane, rotated about z-axis. So distance from z-axis to center of that circle is 2. The point P is at distance sqrt(x^2+y^2) = 1 from z-axis. That’s inside the torus? The torus tube radius is 1, major radius 2. Points inside the torus satisfy (sqrt(x^2+y^2)-2)^2 + z^2 < 1. For P: sqrt(x^2+y^2)=1, so (1-2)^2 + 0 = 1, so on the inner side? Actually equality means on surface? Wait our surface S is exactly where (sqrt(x^2+y^2)-2)^2 + z^2 = 1? Let's verify: from parametrization, we have sqrt(x^2+y^2) = |cosφ+2|. Since cosφ+2 ≥ 1, positive. Then (sqrt(x^2+y^2)-2)^2 + z^2 = (cosφ+2 - 2)^2 + sin²φ = cos²φ + sin²φ =1. So S is indeed the torus surface with major radius 2, minor radius 1.

Interior V: points satisfying (sqrt(x^2+y^2)-2)^2 + z^2 ≤ 1.

At P: (1-2)^2 + 0 = 1, so on the boundary. Inward normal should point toward interior, i.e., direction where (sqrt(x^2+y^2)-2)^2 + z^2 decreases. Since P is on the inner side (closer to z-axis), interior is further away from z-axis? Actually for a given point on torus, interior is inside the tube. The tube cross-section is a circle centered at radius 2 from z-axis. For points where sqrt(x^2+y^2) < 2, the interior direction is toward larger sqrt(x^2+y^2) (i.e., outward from the hole). For P, sqrt(x^2+y^2)=1, so interior direction should increase distance from z-axis. That means moving radially outward from the z-axis. The normal vector we computed n = (-1/√2, -1/√2, 0) points toward decreasing x and y, i.e., toward the z-axis (since at (0.707,0.707,0), moving opposite to that vector gives (1.207,1.207,0) which increases distance). Wait check: n = (-0.707, -0.707, 0). Adding a small step in direction of n: P + ε n = (0.707 - 0.707ε, 0.707 - 0.707ε, 0). This decreases x and y, so distance from z-axis decreases. That’s moving toward z-axis, which is inside the hole, not interior of tube. So n is outward? Actually we need inward normal (pointing into V). Since V is interior of torus, at P (on inner side), interior is in the direction of increasing distance from z-axis (toward the tube center). So the inward normal should be opposite to n we computed? But we computed N = F_θ × F_φ. The orientation depends on order of parameters. Typically outward normal is F_θ × F_φ for a surface given as (θ,φ) with θ first? But we need inward. Let's determine orientation: For a torus parametrization as given, with θ being azimuthal angle, φ being poloidal angle, the usual outward normal (pointing outside the torus) is (cosφ cosθ, cosφ sinθ, sinφ) times (cosφ+2)? Actually our N = (a cosθ cosφ, a sinθ cosφ, a sinφ) = (cosφ+2)(cosθ cosφ, sinθ cosφ, sinφ). That vector points radially outward from the tube's center? At φ=0, N = (a cosθ, a sinθ, 0) = ( (2+cosφ)cosθ, ... ) points away from z-axis? Actually at φ=0, a=3, N = (3 cosθ, 3 sinθ, 0) which points radially outward from z-axis. But at φ=π, a=1, N = (cosθ cosφ, sinθ cosφ, sinφ) with cosφ=-1, sinφ=0 → N = (-cosθ, -sinθ, 0). At P, θ=π/4, N = (-1/√2, -1/√2, 0) which points toward the z-axis (since it's opposite to radial direction (cosθ, sinθ)). So N points inward (toward the hole) at inner side. That suggests N is actually inward normal? But we need inward normal for region V. Since V is interior of torus, inward normal should point into V, i.e., toward the tube interior. At inner side, the tube interior is outward from the hole (increasing distance from z-axis). So inward normal should be opposite to N. Let's check with a point on outer side: φ=0, θ=0, point (3,0,0). Then N = (3*1, 0, 0) = (3,0,0). That points outward from z-axis. But at outer side, interior of torus is toward z-axis (since tube center is at radius 2). So inward normal should point toward z-axis, i.e., negative x direction. But N points in positive x direction, which is outward. So indeed N points outward (away from interior) for the standard orientation (θ,φ). Therefore inward normal is -N/||N||.

Thus at P, inward unit normal n_in = -N/||N|| = -(-1/√2, -1/√2, 0) = (1/√2, 1/√2, 0). Wait that gives (1/√2,1/√2,0) which is exactly P direction? Actually P = (1/√2,1/√2,0) is unit vector radially outward. That would point outward from z-axis, which is toward the interior? At inner side, interior is outward from z-axis. So yes, inward normal at P is radially outward. So n_in = (1/√2, 1/√2, 0).

Thus n1 = 1/√2, n2 = 1/√2.

But note: The problem statement says "Let n = (n1,n2,n3) be the inward unit normal vector at P obtained in I." So we'll use that.

Thus n1 = 1/√2, n2 = 1/√2.

II. Area of portion of S included in region W: W is cylinder x^2+y^2 ≤ 4.

Surface S: torus with major radius 2, minor radius 1. The cylinder radius is 2. Intersection: points on torus satisfying sqrt(x^2+y^2) ≤ 2.

From parametrization: sqrt(x^2+y^2) = |cosφ+2| = cosφ+2 (since cosφ+2 ≥ 1). So condition cosφ+2 ≤ 2 → cosφ ≤ 0 → φ ∈ [π/2, 3π/2] (since φ in [0,2π)).

Thus portion of S in W is φ ∈ [π/2, 3π/2].

Surface area element: dS = ||F_θ × F_φ|| dθ dφ = ||N|| dθ dφ. We computed ||N|| = |a| sqrt(cos²φ+ sin²φ)? Actually N = (a cosθ cosφ, a sinθ cosφ, a sinφ). Its magnitude: sqrt( a² cos²θ cos²φ + a² sin²θ cos²φ + a² sin²φ ) = |a| sqrt( cos²φ(cos²θ+sin²θ) + sin²φ ) = |a| sqrt( cos²φ + sin²φ ) = |a| = |cosφ+2| = cosφ+2 (since positive). So ||N|| = cosφ+2.

Thus area A_S = ∫{θ=0}^{2π} ∫{φ=π/2}^{3π/2} (cosφ+2) dφ dθ = 2π ∫_{π/2}^{3π/2} (cosφ+2) dφ.

Compute integral: ∫ cosφ dφ = sinφ, ∫ 2 dφ = 2φ. Evaluate from π/2 to 3π/2:

sin(3π/2) = -1, sin(π/2)=1 → sinφ difference = -1 - 1 = -2. 2φ: 2*(3π/2) - 2*(π/2) = 3π - π = 2π.

So ∫ = (-2) + 2π = 2π - 2.

Thus A_S = 2π * (2π - 2) = 4π² - 4π.

But we must verify: area element is correct? Actually dS = ||F_θ × F_φ|| dθ dφ = (cosφ+2) dθ dφ. Yes.

Thus A_S = 4π² - 4π.

III. Volume of overlapping region V ∩ W: V is interior of torus (major radius 2, minor radius 1). W is cylinder radius 2. Overlap is intersection of torus interior with cylinder x^2+y^2 ≤ 4.

Since torus interior: (sqrt(x^2+y^2)-2)^2 + z^2 ≤ 1.

We need volume where also x^2+y^2 ≤ 4.

We can compute volume using cylindrical coordinates: ρ = sqrt(x^2+y^2), φ (angle), z.

Torus condition: (ρ - 2)^2 + z^2 ≤ 1.

Cylinder condition: ρ ≤ 2.

Thus region: ρ ≤ 2 and (ρ-2)^2 + z^2 ≤ 1.

Since ρ ≤ 2, ρ-2 ≤ 0, so (ρ-2)^2 = (2-ρ)^2.

Thus condition: (2-ρ)^2 + z^2 ≤ 1.

So region in cylindrical coordinates: 0 ≤ ρ ≤ 2, φ ∈ [0,2π], z satisfying z² ≤ 1 - (2-ρ)^2.

Thus |z| ≤ sqrt(1 - (2-ρ)^2). But note that (2-ρ)^2 ≤ 1 for ρ ∈ [1,3]. Since ρ ≤ 2, we need 2-ρ ≤ 1 → ρ ≥ 1. So ρ ∈ [1,2].

Thus volume V_ov = ∫{ρ=1}^{2} ∫{φ=0}^{2π} ∫_{z=-sqrt(1-(2-ρ)^2)}^{sqrt(...)} ρ dz dφ dρ.

= 2π ∫_{ρ=1}^{2} 2 ρ sqrt(1 - (2-ρ)^2) dρ.

Let u = 2-ρ, then ρ = 2-u, dρ = -du. When ρ=1, u=1; ρ=2, u=0.

Thus V_ov = 2π ∫{u=1}^{0} 2 (2-u) sqrt(1-u^2) (-du) = 4π ∫{0}^{1} (2-u) sqrt(1-u^2) du.

Compute I = ∫ (2-u) sqrt(1-u^2) du from 0 to 1.

Let’s compute analytically or numerically.

I = 2∫ sqrt(1-u^2) du - ∫ u sqrt(1-u^2) du.

First integral: ∫ sqrt(1-u^2) du from 0 to 1 = π/4 (area quarter circle radius 1).

Second integral: ∫ u sqrt(1-u^2) du. Let w = 1-u^2, dw = -2u du → u du = -dw/2. When u=0, w=1; u=1, w=0.

∫ u sqrt(1-u^2) du = ∫ sqrt(w) (-dw/2) = -1/2 * (2/3) w^{3/2} = -1/3 w^{3/2}. Evaluate from u=0 to 1: at u=1, w=0 → 0; at u=0, w=1 → -1/3. So ∫_{0}^{1} u sqrt(1-u^2) du = [0 - (-1/3)] = 1/3.

Thus I = 2*(π/4) - 1/3 = π/2 - 1/3.

Thus V_ov = 4π * (π/2 - 1/3) = 4π*(π/2) - 4π*(1/3) = 2π² - (4π/3).

So V_ov = 2π² - 4π/3.

IV. Curve C on S defined by θ = φ. So parametrization: c(t) with t = θ = φ.

Thus x(t) = (cos t + 2) cos t, y(t) = (cos t + 2) sin t, z(t) = sin t.

We need curvature at point Q = (0,2,1). Find t such that c(t) = Q.

We have x = (cos t + 2) cos t = cos t cos t + 2 cos t = cos² t + 2 cos t. y = (cos t + 2) sin t = cos t sin t + 2 sin t. z = sin t.

Set Q = (0,2,1). So z = sin t = 1 → t = π/2 (since t in [0,2π)). Check: sin(π/2)=1, cos(π/2)=0.

Then x = cos²(π/2) + 2 cos(π/2) = 0 + 0 = 0. y = cos(π/2) sin(π/2) + 2 sin(π/2) = 0 + 2*1 = 2. Yes.

So t = π/2.

Now compute curvature κ(t) = || c' × c'' || / ||c'||^3.

First compute c'(t) = derivative.

Let’s compute symbolically.

Define: x(t) = (cos t + 2) cos t = cos² t + 2 cos t. y(t) = (cos t + 2) sin t = cos t sin t + 2 sin t. z(t) = sin t.

Better to compute derivatives:

x' = derivative of cos² t is -2 cos t sin t, derivative of 2 cos t is -2 sin t. So x' = -2 cos t sin t - 2 sin t = -2 sin t (cos t + 1).

y' = derivative of cos t sin t = cos² t - sin² t? Wait derivative of cos t sin t = -sin t sin t + cos t cos t = cos² t - sin² t = cos 2t. Actually derivative of cos t sin t = cos^2 t - sin^2 t = cos 2t. And derivative of 2 sin t = 2 cos t. So y' = cos 2t + 2 cos t.

z' = cos t.

Second derivatives:

x'' = derivative of -2 sin t (cos t + 1) = -2[ cos t (cos t + 1) + sin t (-sin t) ]? Use product rule: derivative of sin t is cos t, derivative of (cos t+1) is -sin t.

So x'' = -2[ cos t (cos t + 1) + sin t (-sin t) ] = -2[ cos² t + cos t - sin² t ] = -2[ (cos² t - sin² t) + cos t ] = -2[ cos 2t + cos t ].

y'' = derivative of cos 2t + 2 cos t = -2 sin 2t - 2 sin t = -2 sin 2t - 2 sin t.

z'' = - sin t.

Now evaluate at t = π/2:

sin(π/2)=1, cos(π/2)=0.

Compute c'(π/2):

x' = -2 sin t (cos t + 1) = -21(0+1) = -2. y' = cos 2t + 2 cos t = cos π + 2*0 = -1 + 0 = -1. z' = cos t = 0.

So c' = (-2, -1, 0). ||c'|| = sqrt(4+1) = √5.

c''(π/2):

x'' = -2[ cos 2t + cos t ] = -2[ cos π + cos(π/2) ] = -2[ -1 + 0 ] = -2*(-1) = 2. y'' = -2 sin 2t - 2 sin t = -2 sin π - 2*1 = 0 - 2 = -2. z'' = - sin t = -1.

So c'' = (2, -2, -1).

Now compute cross product c' × c'':

c' = (-2, -1, 0), c'' = (2, -2, -1).

Cross product = determinant: i component: (-1)(-1) - (0)(-2) = 1 - 0 = 1. j component: -[ (-2)(-1) - (0)(2) ] = -[ 2 - 0 ] = -2. k component: (-2)(-2) - (-1)(2) = 4 - (-2) = 4+2 = 6.

Thus c' × c'' = (1, -2, 6). Norm = sqrt(1 + 4 + 36) = sqrt(41).

Thus κ = ||c' × c''|| / ||c'||^3 = sqrt(41) / (√5)^3 = sqrt(41) / (5√5) = sqrt(41) / (5√5) = sqrt(41)/(5√5). Rationalize? We'll keep as expression.

κ_Q = sqrt(41) / (5√5).

Now compute (n1 + n2)^2: n1 = n2 = 1/√2, so sum = 2/√2 = √2. Square = 2.

Thus (n1+n2)^2 = 2.

Now compute K = 999 * 123 * ( V_ov/(2π) - A_S/(4π) ) * ( (n1+n2)^2 / κ_Q^2 ) + 7.

First compute V_ov/(2π) = (2π² - 4π/3) / (2π) = (2π²)/(2π) - (4π/3)/(2π) = π - (4π/3)(1/(2π)) = π - (4/3)(1/2) = π - 2/3.

Wait compute carefully: (4π/3)/(2π) = (4π/3)(1/(2π)) = (4/3)(1/2) = 2/3. So V_ov/(2π) = π - 2/3.

Next A_S/(4π) = (4π² - 4π)/(4π) = (4π²)/(4π) - (4π)/(4π) = π - 1.

Thus V_ov/(2π) - A_S/(4π) = (π - 2/3) - (π - 1) = π - 2/3 - π + 1 = 1 - 2/3 = 1/3.

So that term = 1/3.

Now (n1+n2)^2 / κ_Q^2 = 2 / ( (sqrt(41)/(5√5))^2 ) = 2 / (41/(25*5)) = 2 / (41/125) = 2 * (125/41) = 250/41.

Thus product inside: (V_ov/(2π) - A_S/(4π)) * ((n1+n2)^2 / κ_Q^2) = (1/3) * (250/41) = 250/(123).

Now multiply by 999 * 123: 999 * 123 * (250/123) = 999 * 250 = 249750.

Then add 7: K = 249750 + 7 = 249757.

Thus K = 249757.

We need to verify calculations with Python to avoid arithmetic errors.

Let's compute numerically. </think>

[THOUGHT] I will compute each part numerically using Python to verify.

1. Inward normal at P: compute N, then inward direction.
2. Area A_S: integrate (cosφ+2) over θ and φ range.
3. Volume V_ov: integrate in cylindrical coordinates.
4. Curvature κ_Q: compute derivatives at t=π/2.
5. Compute K.

Let's code. </think>

[PYTHON] import sympy as sp import numpy as np

I. Inward unit normal at P

theta = sp.pi/4 phi = sp.pi a = sp.cos(phi) + 2 N = sp.Matrix([a*sp.cos(theta)sp.cos(phi), asp.sin(theta)sp.cos(phi), asp.sin(phi)])

Compute outward normal N (from param order theta, phi)

Inward normal = -N/||N|| (since N points outward as reasoned)

N_norm = sp.sqrt(N.dot(N)) n_inward = -N/N_norm print("Inward unit normal vector at P:", n_inward)

Check magnitude

print("Magnitude of n_inward:", sp.simplify(sp.sqrt(n_inward.dot(n_inward))))

Extract components

n1 = sp.simplify(n_inward[0]) n2 = sp.simplify(n_inward[1]) print("n1 =", n1) print("n2 =", n2)

II. Area A_S

theta_sym, phi_sym = sp.symbols('theta phi', real=True) A = sp.integrate(sp.cos(phi_sym)+2, (phi_sym, sp.pi/2, 3sp.pi/2)) A_S = 2sp.pi * A print("Area A_S =", sp.simplify(A_S))

Alternatively compute integral directly:

A_S_direct = sp.integrate(sp.integrate(sp.cos(phi_sym)+2, (phi_sym, sp.pi/2, 3sp.pi/2)), (theta_sym, 0, 2sp.pi)) print("A_S direct =", sp.simplify(A_S_direct))

III. Volume V_ov

rho = sp.symbols('rho', real=True)

Volume integrand: 2*pi * 2 * rho * sqrt(1 - (2-rho)^2)

But we must integrate rho from 1 to 2.

Compute integral I = 2 * int_{rho=1}^{2} rho * sqrt(1 - (2-rho)^2) d rho

Let u = 2 - rho

u = sp.symbols('u') I_expr = sp.integrate(2*(2-u)sp.sqrt(1-u**2), (u, 0, 1)) V_ov = 2sp.pi * I_expr print("Volume V_ov =", sp.simplify(V_ov))

Alternatively compute directly in rho:

V_ov_direct = sp.integrate(2sp.pi2rhosp.sqrt(1-(2-rho)**2), (rho, 1, 2)) print("V_ov direct =", sp.simplify(V_ov_direct))

IV. Curvature κ_Q at Q (t = π/2)

t = sp.symbols('t')

Parametrization of C: θ = φ = t

x = (sp.cos(t) + 2) * sp.cos(t) y = (sp.cos(t) + 2) * sp.sin(t) z = sp.sin(t) c = sp.Matrix([x, y, z]) c_prime = sp.diff(c, t) c_double = sp.diff(c_prime, t)

Evaluate at t = pi/2

t_val = sp.pi/2 c_prime_val = c_prime.subs(t, t_val) c_double_val = c_double.subs(t, t_val)

Compute curvature

cross_prod = c_prime_val.cross(c_double_val) kappa_val = sp.sqrt(cross_prod.dot(cross_prod)) / (c_prime_val.dot(c_prime_val)**(sp.Rational(3,2))) print("c' at Q:", c_prime_val) print("c'' at Q:", c_double_val) print("||c'|| =", sp.sqrt(c_prime_val.dot(c_prime_val))) print("κ_Q =", sp.simplify(kappa_val)) print("κ_Q numeric:", kappa_val.evalf())

Compute (n1+n2)^2

n1_val = sp.simplify(n1) n2_val = sp.simplify(n2) sum_sq = (n1_val + n2_val)**2 print("(n1+n2)^2 =", sum_sq)

Now compute K

V_ov_val = sp.simplify(V_ov) A_S_val = sp.simplify(A_S) kappa_Q_sq = sp.simplify(kappa_val**2)

term1 = V_ov_val/(2sp.pi) - A_S_val/(4sp.pi) print("term1 = V_ov/(2π) - A_S/(4π) =", term1) print("term1 numeric:", term1.evalf())

term2 = sum_sq / kappa_Q_sq print("term2 = (n1+n2)^2 / κ_Q^2 =", term2) print("term2 numeric:", term2.evalf())

K_expr = 999 * 123 * term1 * term2 + 7 K_val = sp.simplify(K_expr) print("K =", K_val) print("K numeric:", K_val.evalf()) print("K as integer:", int(K_val)) [/PYTHON]

[THOUGHT] The Python output shows ( K = 249757 ) as an integer. This is the final computed value.

\boxed{249757}