[THOUGHT] To solve the problem, we need to find the direction vector of line L₁ by computing the cross product of the normal vectors of the given planes. Using this direction vector, we determine the parametric equations of line L₂. We then find the intersection point Q of L₂ with the plane Π. Next, we calculate the foot of the perpendicular R from point P to the plane Π. Finally, we compute the length PQ and the area of triangle PQR to find the value of 4ℓ²A².

1. Direction Vector of L₁: The cross product of the normal vectors (2, 3, 1) and (1, 2, 1) gives the direction vector (1, -1, 1).
2. Parametric Equations of L₂: Through point P(2, -1, 3) with direction vector (1, -1, 1).
3. Intersection Point Q: Solve for t in the plane equation 2x + y - 2z = 6 using the parametric equations of L₂.
4. Foot of Perpendicular R: Use the parametric line from P in the direction of the normal vector (2, 1, -2) of plane Π and solve for the intersection with Π.
5. Length PQ: Compute the distance between P and Q.
6. Area of Triangle PQR: Use the cross product of vectors PQ and PR to find the area.

[PYTHON] import numpy as np

Step 1: Direction vector of L1 (cross product of normals)

n1 = np.array([2, 3, 1]) n2 = np.array([1, 2, 1]) direction_L1 = np.cross(n1, n2)

Step 2: Parametric equations of L2 through P(2, -1, 3)

P = np.array([2, -1, 3]) t = np.linspace(-10, 0, 1000) # Parameter t for L2 L2 = P + t[:, np.newaxis] * direction_L1

Step 3: Find intersection Q of L2 with plane Π: 2x + y - 2z = 6

Solve 2*(2 + t) + (-1 - t) - 2*(3 + t) = 6

t_Q = np.roots([-3, -3 + 4 - 1 - 2]) # Coefficients of the equation -3t -3 = 6 t_Q = t_Q[np.isreal(t_Q)].real[0] Q = P + t_Q * direction_L1

Step 4: Foot of perpendicular R from P to Π

normal_Pi = np.array([2, 1, -2]) s = np.roots([normal_Pi @ normal_Pi, normal_Pi @ (P - 6/np.linalg.norm(normal_Pi))]) s = s[np.isreal(s)].real[0] R = P + s * normal_Pi

Step 5: Length PQ

PQ = Q - P ell_squared = np.dot(PQ, PQ)

Step 6: Area of triangle PQR

PR = R - P cross_product = np.cross(PQ, PR) A_squared = 0.25 * np.dot(cross_product, cross_product)

Calculate 4 * ell^2 * A^2

result = 4 * ell_squared * A_squared print(int(result)) [/PYTHON]

[THOUGHT] The Python code computed the value of (4\ell^2 A^2) as 4. Based on the output, the answer is 4. \boxed{4}