The square of the distance of the image of the point $A(6, 1, 5)$ in the line $\frac{x-1}{3} = \frac{y}{2} = \frac{z-2}{4}$ from the origin 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 angle between the lines $\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}$ and $\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}$ is

The lines $\overrightarrow{r} = (\hat{i} - \hat{j}) + \ell(2\hat{i} + \hat{k})$ and $\overrightarrow{r} = (2\hat{i} - \hat{j}) + m(\hat{i} + \hat{j} - \hat{k})$:

The equation of the straight line passing through the point $(a, b, c)$ and parallel to the $z$-axis is:

If the harmonic conjugate of $P(2, 3, 4)$ with respect to the line segment joining the points $A(3, -2, 2)$ and $B(6, -17, -4)$ is $Q(\alpha, \beta, \gamma)$,then $\alpha + \beta + \gamma =$