Let the image of the point P(0, -5, 0) in the line (x-1)/(2) =...

Let $M_1$ be the foot of the perpendicular from $P(0, -5, 0)$ to the first line $L_1: \dfrac{x-1}{2} = \dfrac{y}{1} = \dfrac{z+1}{-2}$.
A general point on $L_1$ is $M_1(2\lambda + 1, \lambda, -2\lambda - 1)$.
The direction ratios of $PM_1$ are $(2\lambda + 1, \lambda + 5, -2\lambda - 1)$.
Since $PM_1$ is perpendicular to $L_1$, their dot product is zero:
$2(2\lambda + 1) + 1(\lambda + 5) - 2(-2\lambda - 1) = 0$
$4\lambda + 2 + \lambda + 5 + 4\lambda + 2 = 0 \Rightarrow 9\lambda + 9 = 0 \Rightarrow \lambda = -1$
Thus, $M_1$ is $(-1, -1, 1)$.

Let $M_2$ be the foot of the perpendicular from $Q\left(0, \dfrac{-1}{2}, 0\right)$ to the second line $L_2: \dfrac{x-1}{-1} = \dfrac{y+9}{4} = \dfrac{z+1}{1}$.
A general point on $L_2$ is $M_2(-\mu + 1, 4\mu - 9, \mu - 1)$.
The direction ratios of $QM_2$ are \left(-\mu + 1, 4\mu - \dfrac{17}{2}, \mu - 1\right)$.<br>Since$QM_2$is perpendicular to$L_2$, their dot product is zero:<br>$-1(-\mu + 1) + 4\left(4\mu - \dfrac{17}{2}\right) + 1(\mu - 1) = 0$<br>$\mu - 1 + 16\mu - 34 + \mu - 1 = 0 \Rightarrow 18\mu - 36 = 0 \Rightarrow \mu = 2$<br>Thus,$M_2$is$(-1, -1, 1)$.<br><br>Since$M_1$and$M_2$are the same point$M(-1, -1, 1)$, the diagonals$PR$and$QS$of the quadrilateral$PQRS$bisect each other at$M$. Therefore,$PQRS$is a parallelogram.<br><br>The area of the parallelogram is given by$\dfrac{1}{2} |\overrightarrow{PR} \times \overrightarrow{QS}|$.<br>We have:<br>$\overrightarrow{PR} = 2\overrightarrow{PM} = 2(-1 - 0, -1 - (-5), 1 - 0) = (-2, 8, 2)$<br>$\overrightarrow{QS} = 2\overrightarrow{QM} = 2\left(-1 - 0, -1 - \left(\dfrac{-1}{2}\right), 1 - 0\right) = (-2, -1, 2)$<br><br>Now, compute the cross product$\overrightarrow{PR} \times \overrightarrow{QS}$:<br>$\overrightarrow{PR} \times \overrightarrow{QS} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 8 & 2 \\ -2 & -1 & 2 \end{vmatrix} = \hat{i}(16 - (-2)) - \hat{j}(-4 - (-4)) + \hat{k}(2 - (-16)) = 18\hat{i} + 18\hat{k}$<br><br>The area of the parallelogram is:<br>$\text{Area} = \dfrac{1}{2} |18\hat{i} + 18\hat{k}| = \dfrac{1}{2} \sqrt{18^2 + 18^2} = \dfrac{18\sqrt{2}}{2} = 9\sqrt{2}$<br><br>The square of the area is$(9\sqrt{2})^2 = 162$.<br><br>Answer:$162