Problem Analysis #13

[THOUGHT] To find the value of (4\ell^2 A^2), we need to go through several steps:

1. Find the direction vector of (L_1) by taking the cross product of the normal vectors of the given planes.
2. Determine the parametric equations of (L_2) using the direction vector from (L_1) and the given point (P).
3. Find the coordinates of point (Q) where (L_2) intersects the plane (\Pi).
4. Calculate the length (\ell) of (PQ).
5. Find the coordinates of point (R), the foot of the perpendicular from (P) to the plane (\Pi).
6. Calculate the area (A) of (\triangle PQR).
7. Compute (4\ell^2 A^2).

Let's start with step 1.

The normal vectors of the planes (2x + 3y + z = 4) and (x + 2y + z = 5) are (\mathbf{n}_1 = \langle 2, 3, 1 \rangle) and (\mathbf{n}_2 = \langle 1, 2, 1 \rangle).

The direction vector (\mathbf{d}) of (L_1) is the cross product (\mathbf{n}_1 \times \mathbf{n}_2): [ \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & 3 & 1 \ 1 & 2 & 1 \end{vmatrix} = \mathbf{i}(3 \cdot 1 - 1 \cdot 2) - \mathbf{j}(2 \cdot 1 - 1 \cdot 1) + \mathbf{k}(2 \cdot 2 - 3 \cdot 1) = \langle 1, -1, 1 \rangle ]

For step 2, the parametric equations of (L_2) passing through (P(2, -1, 3)) and parallel to (\mathbf{d}) are: [ x = 2 + t, \quad y = -1 - t, \quad z = 3 + t ]

For step 3, substitute these parametric equations into the plane equation (2x + y - 2z = 6): [ 2(2 + t) + (-1 - t) - 2(3 + t) = 6 ] [ 4 + 2t - 1 - t - 6 - 2t = 6 ] [ -3 - t = 6 ] [ t = -9 ] Substitute (t = -9) back into the parametric equations to find (Q): [ x = 2 - 9 = -7, \quad y = -1 + 9 = 8, \quad z = 3 - 9 = -6 ] So, (Q = (-7, 8, -6)).

For step 4, calculate the length (\ell) of (PQ): [ PQ = \sqrt{(-7 - 2)^2 + (8 + 1)^2 + (-6 - 3)^2} = \sqrt{(-9)^2 + 9^2 + (-9)^2} = \sqrt{81 + 81 + 81} = \sqrt{243} = 9\sqrt{3} ] Thus, (\ell = 9\sqrt{3}).

For step 5, find the coordinates of (R), the foot of the perpendicular from (P(2, -1, 3)) to the plane (2x + y - 2z = 6). The formula for the foot of the perpendicular from a point ((x_0, y_0, z_0)) to a plane (Ax + By + Cz + D = 0) is: [ R = \left( x_0 - \frac{A(Ax_0 + By_0 + Cz_0 + D)}{A^2 + B^2 + C^2}, y_0 - \frac{B(Ax_0 + By_0 + Cz_0 + D)}{A^2 + B^2 + C^2}, z_0 - \frac{C(Ax_0 + By_0 + Cz_0 + D)}{A^2 + B^2 + C^2} \right) ] For the plane (2x + y - 2z - 6 = 0) and point (P(2, -1, 3)): [ A = 2, \quad B = 1, \quad C = -2, \quad D = -6 ] [ 2(2) + 1(-1) - 2(3) - 6 = 4 - 1 - 6 - 6 = -9 ] [ A^2 + B^2 + C^2 = 2^2 + 1^2 + (-2)^2 = 4 + 1 + 4 = 9 ] [ R = \left( 2 - \frac{2(-9)}{9}, -1 - \frac{1(-9)}{9}, 3 - \frac{-2(-9)}{9} \right) = \left( 2 + 2, -1 + 1, 3 - 2 \right) = (4, 0, 1) ]

For step 6, calculate the area (A) of (\triangle PQR). We use the formula for the area of a triangle given by three points ((x_1, y_1, z_1)), ((x_2, y_2, z_2)), and ((x_3, y_3, z_3)): [ A = \frac{1}{2} \left| \mathbf{PQ} \times \mathbf{PR} \right| ] First, find vectors (\mathbf{PQ}) and (\mathbf{PR}): [ \mathbf{PQ} = \langle -7 - 2, 8 + 1, -6 - 3 \rangle = \langle -9, 9, -9 \rangle ] [ \mathbf{PR} = \langle 4 - 2, 0 + 1, 1 - 3 \rangle = \langle 2, 1, -2 \rangle ] Next, calculate the cross product (\mathbf{PQ} \times \mathbf{PR}): [ \mathbf{PQ} \times \mathbf{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -9 & 9 & -9 \ 2 & 1 & -2 \end{vmatrix} = \mathbf{i}(9 \cdot -2 - (-9) \cdot 1) - \mathbf{j}(-9 \cdot -2 - (-9) \cdot 2) + \mathbf{k}(-9 \cdot 1 - 9 \cdot 2) ] [ = \mathbf{i}(-18 + 9) - \mathbf{j}(18 - 18) + \mathbf{k}(-9 - 18) = \langle -9, 0, -27 \rangle ] Find the magnitude of this vector: [ \left| \mathbf{PQ} \times \mathbf{PR} \right| = \sqrt{(-9)^2 + 0^2 + (-27)^2} = \sqrt{81 + 0 + 729} = \sqrt{810} = 9\sqrt{10} ] Thus, the area (A) is: [ A = \frac{1}{2} \cdot 9\sqrt{10} = \frac{9\sqrt{10}}{2} ]

For step 7, compute (4\ell^2 A^2): [ 4\ell^2 A^2 = 4 \left( 9\sqrt{3} \right)^2 \left( \frac{9\sqrt{10}}{2} \right)^2 = 4 \cdot 243 \cdot \frac{810}{4} = 243 \cdot 810 = 198270 ]

The final answer is: [ \boxed{198270} ]