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:

$$\boxed{198270}$$