The problem asks for the value of $4\ell^2 A^2$, where $\ell$ is the length of segment $PQ$ and $A$ is the area of triangle $\triangle PQR$.

First, I will determine the direction of line $L_1$ by taking the cross product of the normal vectors of the two planes defining it. Then, I will define line $L_2$ passing through $P(2, -1, 3)$ with this direction. Next, I will find the intersection point $Q$ of line $L_2$ and the plane $\Pi: 2x + y - 2z = 6$. I will also find the point $R$, which is the projection of $P$ onto plane $\Pi$. Since $PR$ is perpendicular to the plane $\Pi$ and $Q$ lies in $\Pi$, the triangle $\triangle PQR$ is a right-angled triangle with the right angle at $R$. Finally, I will calculate $\ell = PQ$, $h = PR$, and the area $A = \frac{1}{2} \times PR \times RQ$, and compute the required value.

[PYTHON] import numpy as np

Define normal vectors for the planes defining L1

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

Direction vector of L1 (and L2)

d = np.cross(n1, n2)

print(f"Direction vector d: {d}")

Point P

P = np.array([2, -1, 3])

Plane Pi: 2x + y - 2z = 6

Normal vector n_pi and constant k_pi

n_pi = np.array([2, 1, -2]) k_pi = 6

Find Q: Intersection of L2 and Pi

L2: r = P + t * d

Substitute into plane equation: n_pi . (P + t * d) = k_pi

n_pi . P + t * (n_pi . d) = k_pi

t = (k_pi - n_pi . P) / (n_pi . d)

numerator = k_pi - np.dot(n_pi, P) denominator = np.dot(n_pi, d)

if denominator == 0: print("Line is parallel to the plane.") else: t_Q = numerator / denominator Q = P + t_Q * d

# Calculate length l = PQ
l_vec = Q - P
l = np.linalg.norm(l_vec)
l_sq = np.dot(l_vec, l_vec) # l^2

# Find R: Projection of P onto Pi
# Line PR is parallel to n_pi
# R = P + s * n_pi
# Substitute into plane equation: n_pi . (P + s * n_pi) = k_pi
# s = (k_pi - n_pi . P) / (n_pi . n_pi)

s_R = (k_pi - np.dot(n_pi, P)) / np.dot(n_pi, n_pi)
R = P + s_R * n_pi

# Calculate height h = PR
h_vec = R - P
h = np.linalg.norm(h_vec)
h_sq = np.dot(h_vec, h_vec) # h^2

# Calculate RQ
rq_vec = Q - R
rq_sq = np.dot(rq_vec, rq_vec)

# Verify right triangle
# dot_prod = np.dot(h_vec, rq_vec)
# print(f"Dot product PR . RQ: {dot_prod}") # Should be 0

# Area A = 0.5 * PR * RQ
# A^2 = 0.25 * PR^2 * RQ^2
A_sq = 0.25 * h_sq * rq_sq

result = 4 * l_sq * A_sq

print(f"Direction d: {d}")
print(f"P: {P}")
print(f"Q: {Q}")
print(f"R: {R}")
print(f"l^2: {l_sq}")
print(f"h^2 (PR^2): {h_sq}")
print(f"RQ^2: {rq_sq}")
print(f"A^2: {A_sq}")
print(f"Result: {result}")

[/PYTHON]

The direction vector of $L_1$ is the cross product of the normal vectors of the two planes: [ \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & 3 & 1 \ 1 & 2 & 1 \end{vmatrix} = (3(1) - 1(2))\mathbf{i} - (2(1) - 1(1))\mathbf{j} + (2(2) - 3(1))\mathbf{k} = \mathbf{i} - \mathbf{j} + \mathbf{k} = (1, -1, 1). ] The line $L_2$ passes through $P(2, -1, 3)$ and has direction vector $\mathbf{d} = (1, -1, 1)$. Its parametric equations are $x = 2 + t$, $y = -1 - t$, $z = 3 + t$. To find the intersection point $Q$ with the plane $\Pi: 2x + y - 2z = 6$, we substitute the parametric equations into the plane equation: [ 2(2 + t) + (-1 - t) - 2(3 + t) = 6 ] [ 4 + 2t - 1 - t - 6 - 2t = 6 ] [ -3 - t = 6 \implies t = -9. ] The point $Q$ corresponds to $t = -9$: [ Q = (2 - 9, -1 - (-9), 3 - 9) = (-7, 8, -6). ] The length squared of $PQ$ is $\ell^2 = (-7 - 2)^2 + (8 - (-1))^2 + (-6 - 3)^2 = (-9)^2 + 9^2 + (-9)^2 = 81 + 81 + 81 = 243$.

$R$ is the foot of the perpendicular from $P$ to the plane $\Pi$. The distance $PR$ is given by the formula for the distance from a point to a plane: [ PR = \frac{|2(2) + 1(-1) - 2(3) - 6|}{\sqrt{2^2 + 1^2 + (-2)^2}} = \frac{|4 - 1 - 6 - 6|}{\sqrt{9}} = \frac{|-9|}{3} = 3. ] So $PR^2 = 9$. Since $PR \perp \Pi$ and $Q$ lies in $\Pi$, $\triangle PQR$ is a right triangle with hypotenuse $PQ$. By the Pythagorean theorem, $RQ^2 = PQ^2 - PR^2 = 243 - 9 = 234$. The area of $\triangle PQR$ is $A = \frac{1}{2} \cdot PR \cdot RQ$. Thus, $A^2 = \frac{1}{4} \cdot PR^2 \cdot RQ^2 = \frac{1}{4} \cdot 9 \cdot 234 = \frac{2106}{4} = 526.5$. We need to find $4\ell^2 A^2$: [ 4\ell^2 A^2 = 4 \cdot 243 \cdot 526.5 = 243 \cdot 2106 = 511758. ]

\boxed{511758}