[THOUGHT] We need to compute 4 * ℓ^2 * A^2, where ℓ = length PQ, and A = area of triangle PQR.

Given:

• L1: intersection of planes 2x+3y+z=4 and x+2y+z=5.
• L2: line through P(2,-1,3) parallel to L1.
• Plane Π: 2x+y-2z=6.
• Q = intersection of L2 with Π.
• R = foot of perpendicular from P to Π.
• ℓ = distance PQ.
• A = area of triangle PQR.

We need 4 ℓ^2 A^2.

Steps:

1. Find direction vector of L1: cross product of normals of the two planes. Plane1: 2x+3y+z=4 → normal n1 = (2,3,1) Plane2: x+2y+z=5 → normal n2 = (1,2,1) Direction d = n1 × n2 = cross product.
2. L2: through P with direction d.
3. Find Q: intersection of L2 with Π. Parametric eq: P + t d. Plug into plane equation to solve t.
4. Compute ℓ = distance between P and Q = |t| * |d| (or compute directly).
5. Find R: foot of perpendicular from P to Π. Use formula: project P onto plane along normal of Π. Plane Π: 2x+y-2z=6, normal n = (2,1,-2). Find point R such that vector PR is parallel to n and R lies on plane. Let R = P + s n. Plug into plane equation to solve s.
6. Compute area A of triangle PQR. We have coordinates of P, Q, R. Area = 0.5 * |(Q-P) × (R-P)|.
7. Compute ℓ^2 and A^2, then 4 ℓ^2 A^2.

We'll compute exactly using sympy for rational numbers.

Let's implement. [/THOUGHT] [PYTHON] import sympy as sp

Define symbols

t, s = sp.symbols('t s')

Points and vectors

P = sp.Point(2, -1, 3)

Normals

n1 = sp.Matrix([2, 3, 1]) n2 = sp.Matrix([1, 2, 1])

Direction of L1 (cross product)

d = n1.cross(n2) print("Direction vector d:", d)

Parametric line L2: P + t*d

x, y, z = sp.symbols('x y z')

Coordinates as functions of t

x_expr = P.x + t * d[0] y_expr = P.y + t * d[1] z_expr = P.z + t * d[2]

Plane Π: 2x + y - 2z = 6

plane_eq = 2x + y - 2z - 6

Substitute parametric into plane equation

sub_eq = plane_eq.subs({x: x_expr, y: y_expr, z: z_expr}) print("Equation in t:", sub_eq)

Solve for t

t_sol = sp.solve(sub_eq, t) print("t solution:", t_sol) t_val = t_sol[0]

Coordinates of Q

Q = sp.Point(x_expr.subs(t, t_val), y_expr.subs(t, t_val), z_expr.subs(t, t_val)) print("Q:", Q)

Distance ℓ = PQ

PQ_vec = sp.Matrix(Q - P) ell_sq = PQ_vec.dot(PQ_vec) # squared length print("ℓ^2:", ell_sq)

Find R: foot of perpendicular from P to plane Π

Plane normal n = (2,1,-2)

n = sp.Matrix([2, 1, -2])

Parametric: R = P + s * n

R = sp.Point(P.x + s * n[0], P.y + s * n[1], P.z + s * n[2])

R lies on plane: 2Rx + Ry - 2Rz = 6

plane_R = 2R.x + R.y - 2R.z - 6 s_sol = sp.solve(plane_R, s) print("s solution:", s_sol) s_val = s_sol[0] R = sp.Point(P.x + s_val * n[0], P.y + s_val * n[1], P.z + s_val * n[2]) print("R:", R)

Area of triangle PQR: 0.5 * |(Q-P) × (R-P)|

QP = sp.Matrix(Q - P) RP = sp.Matrix(R - P) cross = QP.cross(RP) cross_norm_sq = cross.dot(cross) area_sq = (cross_norm_sq) / 4 # because area = 0.5 * |cross|, so area^2 = (1/4) * |cross|^2 print("Area^2:", area_sq)

Compute 4 * ℓ^2 * A^2

result = 4 * ell_sq * area_sq print("4 ℓ^2 A^2:", result)

Simplify if needed

result_simplified = sp.simplify(result) print("Simplified result:", result_simplified) [/PYTHON]

[THOUGHT] The Python output gives ℓ^2 = 243, area^2 = 1053/2, and 4 ℓ^2 A^2 = 511758. This is the exact integer result we need. [/THOUGHT] \boxed{511758}