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

The coordinates of the foot of the perpendicular from the point $(0,2,3)$ on the line $\frac{x+3}{5}=\frac{y+1}{2}=\frac{z+4}{3}$ are

Let $P(\alpha, \beta, \gamma)$ be the image of the point $Q(3, -3, 1)$ in the line $\frac{x-0}{1} = \frac{y-3}{1} = \frac{z-1}{-1}$ and $R$ be the point $(2, 5, -1)$. If the area of the triangle $PQR$ is $\lambda$ and $\lambda^2 = 14K$,then $K$ is equal to:

Find the shortest distance between lines $\vec{r}=6 \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})$ and $\vec{r}=-4 \hat{i}-\hat{k}+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k})$. (in $units$)

$A$ line with direction ratios $2, 1, 2$ meets the lines $x = y + 2 = z$ and $x + 2 = 2y = 2z$ at points $P$ and $Q$ respectively. If the length of the perpendicular from the point $(1, 2, 12)$ to the line $PQ$ is $l$,then $l^2$ is

If the foot of the perpendicular drawn from $(1, 9, 7)$ to the line passing through the point $(3, 2, 1)$ and parallel to the planes $x+2y+z=0$ and $3y-z=3$ is $(\alpha, \beta, \gamma)$,then $\alpha+\beta+\gamma$ is equal to