The plane containing the line $\frac{x - 3}{2} = \frac{y + 2}{-1} = \frac{z - 1}{3}$ and also containing its projection on the plane $2x + 3y - z = 5$ contains which one of the following points?

If the distance of the point $2 \hat{i} + 3 \hat{j} + \lambda \hat{k}$ from the plane $\vec{r} \cdot (3 \hat{i} + 2 \hat{j} + 6 \hat{k}) = 13$ is $5$ units,then $\lambda =$

The equation of the plane containing the line $\frac{x+1}{2}=\frac{y+2}{1}=\frac{z-2}{3}$ and the point $(1,-1,3)$ is

The line $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ is parallel to the plane

The angle between the line of intersection of the two planes $r \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=5$ and $r \cdot(3 \hat{i}+3 \hat{j}-5 \hat{k})=3$,and the line $r=3 \hat{i}+2 \hat{j}+\hat{k}+t(5 \hat{i}+5 \hat{j}-7 \hat{k})$ is

Let a line $L$ passing through the point $P(1, 1, 1)$ 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 $yz$-plane at the point $Q$. Another line parallel to $L$ and passing through the point $S(1, 0, -1)$ intersects the $yz$-plane at the point $R$. Then the square of the area of the parallelogram $PQRS$ is equal to . . . . . . .