Let a line $L_1$ pass through the origin and be perpendicular to the lines

$\mathrm{L}_2: \overrightarrow{\mathrm{r}}=(3+\mathrm{t}) \hat{i}+(2 \mathrm{t}-1) \hat{j}+(2 \mathrm{t}+4) \hat{k}$ and

$\mathrm{L}_3: \overrightarrow{\mathrm{r}}=(3+2 \mathrm{~s}) \hat{i}+(3+2 \mathrm{~s}) \hat{j}+(2+\mathrm{s}) \hat{k}, \mathrm{t}, \mathrm{s} \in \mathbf{R}$.

If $(a, b, c), a \in \mathbf{Z}$, is the point on $\mathrm{L}_3$ at a distance of $\sqrt{17}$ from the point of intersection of $\mathrm{L}_1$ and $\mathrm{L}_2$, then $(\mathrm{a}+\mathrm{b}+\mathrm{c})^2$ is equal to $\_\_\_\_$ .

Let the image of the point $\mathrm{P}(0,-5,0)$ in the line $\frac{x-1}{2}=\frac{y}{1}=\frac{z+1}{-2}$ be the point R and the image of the point $\mathrm{Q}\left(0, \frac{-1}{2}, 0\right)$ in the line $\frac{x-1}{-1}=\frac{y+9}{4}=\frac{z+1}{1}$ be the point S . Then the square of the area of the parallelogram PQRS is $\_\_\_\_$ .

Let a line L passing through the point $\mathrm{P}(1,1,1)$ be perpendicular to the lines $\frac{x-4}{4}=\frac{y-1}{1}=\frac{z-1}{1}$ and $\frac{x-17}{1}=\frac{y-71}{1}=\frac{z}{0}$. Let the line L intersect the $y z-$ plane at the point Q . Another line parallel to L and passing through the point $\mathrm{S}(1,0,-1)$ intersects the $y z$-plane at the point R . Then the square of the area of the parallelogram PQRS is equal to $\_\_\_\_$ .

If the image of the point $\mathrm{P}(a, 2, a)$ in the line $\frac{x}{2}=\frac{y+a}{1}=\frac{z}{1}$ is Q and the image

of Q in the line $\frac{x-2 b}{2}=\frac{y-a}{1}=\frac{z+2 b}{-5}$ is P , then $a+b$ is equal to $\_\_\_\_$ .