The line passing through the points $(5, 1, a)$ and $(3, b, 1)$ crosses the $yz-$ plane at the point $(0, \frac{17}{2}, -\frac{13}{2})$. Then:

If $P$ is a point lying on the line passing through the point $A(\hat{i}-\hat{j}+3 \hat{k})$ and parallel to the vector $2 \hat{i}+\hat{j}-2 \hat{k}$ such that $|AP|=18$,then a position vector of $P$ is

Let the image of the point $P(0, -5, 0)$ in the line $\frac{x-1}{2} = \frac{y}{1} = \frac{z+1}{-2}$ be the point $R$ and the image of the point $Q(0, -1/2, 0)$ 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 . . . . . . .

One vertex of a rectangular parallelepiped is at the origin $O$ and the lengths of its edges along $x, y$ and $z$ axes are $3, 4$ and $5$ units respectively. Let $P$ be the vertex $(3, 4, 5)$. Then the shortest distance between the diagonal $OP$ and an edge parallel to the $z$-axis,not passing through $O$ or $P$,is:

Let the line $L$ intersect the lines $x-2=-y=z-1$ and $2(x+1)=2(y-1)=z+1$,and be parallel to the line $\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}$. Then which of the following points lies on $L$?

If lines $\frac{x-3}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-1}{1}=\frac{6-z}{5}$ are perpendicular to each other,then $k=$ $\qquad$ .