If lines $\frac{x - 3}{2} = \frac{y + 1}{-3} = \frac{z + a}{p}$ and $\frac{x + 2}{2} = \frac{y - 4}{4} = \frac{z + 5}{2}$ are perpendicular coplanar lines,then the value of $a + p$ is

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect,then the value of $k$ is

The straight line $\frac{x-3}{3}=\frac{y-2}{1}=\frac{z-1}{0}$ is

If the foot of the perpendicular drawn from the point $A(1, 0, 3)$ on a line passing through $B(\alpha, 7, 1)$ is $P\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)$,then $\alpha$ is equal to:

Let $P$ be the foot of the perpendicular from the point $A(1, 2, 2)$ on the line $L: \frac{x-1}{1} = \frac{y+1}{-1} = \frac{z-2}{2}$. Let the line $\overrightarrow{r} = (-\hat{i} + \hat{j} - 2\hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k})$,$\lambda \in R$,intersect the line $L$ at $Q$. Then $2(PQ)^2$ is equal to:

The lines whose direction cosines satisfy the equations $al + bm + cn = 0$ and $fmn + gnl + hlm = 0$ will be perpendicular if...