Square of Area of Parallelogram Formed by Image Points

To find the square of the area of the parallelogram $PQRS$, we first calculate the coordinates of the image points $R$ and $S$.

Step 1: Find the image point $R$ of $P(0, -5, 0)$ in line $L_1$

The line $L_1$ is given by: $\frac{x - 1}{2} = \frac{y}{1} = \frac{z + 1}{-2} = \lambda$

Any general point $M_1$ on line $L_1$ has coordinates: $M_1 = (2\lambda + 1, \lambda, -2\lambda - 1)$

Let $M_1$ be the foot of the perpendicular from $P(0, -5, 0)$ to line $L_1$. The direction vector of $L_1$ is $\vec{d_1} = 2\hat{i} + \hat{j} - 2\hat{k}$, and the vector $\vec{PM_1}$ is given by: $\vec{PM_1} = (2\lambda + 1 - 0)\hat{i} + (\lambda - (-5))\hat{j} + (-2\lambda - 1 - 0)\hat{k} = (2\lambda + 1)\hat{i} + (\lambda + 5)\hat{j} + (-2\lambda - 1)\hat{k}$

Since $\vec{PM_1} \perp \vec{d_1}$, their dot product is zero: $\vec{PM_1} \cdot \vec{d_1} = 0$ $2(2\lambda + 1) + 1(\lambda + 5) - 2(-2\lambda - 1) = 0$ $4\lambda + 2 + \lambda + 5 + 4\lambda + 2 = 0$ $9\lambda + 9 = 0 \implies \lambda = -1$

Substituting $\lambda = -1$, the foot of the perpendicular $M_1$ is: $M_1 = (-1, -1, 1)$

Since $M_1$ is the midpoint of segment $PR$, the coordinates of the image point $R(x_R, y_R, z_R)$ are: $R = 2M_1 - P = 2(-1, -1, 1) - (0, -5, 0) = (-2, 3, 2)$

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Step 2: Find the image point $S$ of $Q\left(0, -\frac{1}{2}, 0\right)$ in line $L_2$

The line $L_2$ is given by: $\frac{x - 1}{-1} = \frac{y + 9}{4} = \frac{z + 1}{1} = \mu$

Any general point $M_2$ on line $L_2$ has coordinates: $M_2 = (-\mu + 1, 4\mu - 9, \mu - 1)$

Let $M_2$ be the foot of the perpendicular from $Q\left(0, -\frac{1}{2}, 0\right)$ to line $L_2$. The direction vector of $L_2$ is $\vec{d_2} = -\hat{i} + 4\hat{j} + \hat{k}$, and the vector $\vec{QM_2}$ is: $\vec{QM_2} = (-\mu + 1)\hat{i} + \left(4\mu - 9 + \frac{1}{2}\right)\hat{j} + (\mu - 1)\hat{k} = (-\mu + 1)\hat{i} + \left(4\mu - \frac{17}{2}\right)\hat{j} + (\mu - 1)\hat{k}$

Since $\vec{QM_2} \perp \vec{d_2}$, their dot product is zero: $\vec{QM_2} \cdot \vec{d_2} = 0$ $-1(-\mu + 1) + 4\left(4\mu - \frac{17}{2}\right) + 1(\mu - 1) = 0$ $\mu - 1 + 16\mu - 34 + \mu - 1 = 0$ $18\mu - 36 = 0 \implies \mu = 2$

Substituting $\mu = 2$, the foot of the perpendicular $M_2$ is: $M_2 = (-1, -1, 1)$

Since $M_2$ is the midpoint of segment $QS$, the coordinates of the image point $S(x_S, y_S, z_S)$ are: $S = 2M_2 - Q = 2(-1, -1, 1) - \left(0, -\frac{1}{2}, 0\right) = \left(-2, -\frac{3}{2}, 2\right)$

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Step 3: Area of the Parallelogram $PQRS$

The adjacent side vectors of the parallelogram $PQRS$ are: $\vec{PQ} = Q - P = \left(0, -\frac{1}{2} - (-5), 0\right) = \left(0, \frac{9}{2}, 0\right)$ $\vec{PR} = R - P = (-2 - 0, 3 - (-5), 2 - 0) = (-2, 8, 2)$

The area vector is given by the cross product $\vec{PQ} \times \vec{PR}$: $\vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & \frac{9}{2} & 0 \\ -2 & 8 & 2 \end{vmatrix}$ $= \hat{i}\left(\frac{9}{2} \cdot 2 - 0\right) - \hat{j}(0 - 0) + \hat{k}\left(0 - \left(\frac{9}{2}\right)(-2)\right)$ $= 9\hat{i} + 9\hat{k}$

The square of the area of the parallelogram $PQRS$ is: $\text{Area}^2 = |\vec{PQ} \times \vec{PR}|^2 = 9^2 + 0^2 + 9^2 = 81 + 81 = 162$