Let $P$ be the foot of the perpendicular from the point $Q(10,-3,-1)$ on the line $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z+1}{-2}$. Then the area of the right-angled triangle $PQR$,where $R$ is the point $(3,-2,1)$,is

Let the line $L_{1}$ be parallel to the vector $-3\hat{i}+2\hat{j}+4\hat{k}$ and pass through the point $(2, 6, 7)$, and the line $L_{2}$ be parallel to the vector $2\hat{i}+\hat{j}+3\hat{k}$ and pass through the point $(4, 3, 5)$. If the line $L_{3}$ is parallel to the vector $-3\hat{i}+5\hat{j}+16\hat{k}$ and intersects the lines $L_{1}$ and $L_{2}$ at the points $C$ and $D$, respectively, then $|\overrightarrow{CD}|^2$ is equal to:

The distance of the point $-\hat{i} + 2\hat{j} + 6\hat{k}$ from the straight line that passes through the point $2\hat{i} + 3\hat{j} - 4\hat{k}$ and is parallel to the vector $6\hat{i} + 3\hat{j} - 4\hat{k}$ is

The point of intersection of the line joining the points $(3, 4, 1)$ and $(5, 1, 6)$ and the $xy$-plane is

The vector equation of the line $\frac{x+3}{2}=\frac{2y-3}{5}; z=-1$ is

Lines $\frac{x-5}{7}=\frac{y-5}{k}=\frac{z-2}{1}$ and $\frac{x}{1}=\frac{y-3}{2}=\frac{z+1}{3}$ are perpendicular to each other,then the value of $k=$ . . . . . . .