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

If the Cartesian equation of a line is $6x-2=3y+1=2z-2$,then the vector equation of the line is

The coordinates of the points on the line $\frac{x+2}{1}=\frac{y-1}{2}=\frac{z+1}{-2}$ at a distance of $12 \text{ units}$ from the point $A(-2, 1, -1)$ are

The distance of the point $({x_1}, {y_1}, {z_1})$ from the line $\frac{{x - {x_2}}}{l} = \frac{{y - {y_2}}}{m} = \frac{{z - {z_2}}}{n}$,where $l, m, n$ are the direction cosines of the line,is given by:

Let a triangle $PQR$ be such that $P$ and $Q$ lie on the line $\frac{x+3}{8} = \frac{y-4}{2} = \frac{z+1}{2}$ and are at a distance of $6$ units from $R(1, 2, 3)$. If $(\alpha, \beta, \gamma)$ is the centroid of $\triangle PQR$, then $\alpha + \beta + \gamma$ is equal to :

Find the shortest distance between the lines $l_{1}$ and $l_{2}$ whose vector equations are
$\vec{r}=\hat{i}+\hat{j}+\lambda(2 \hat{i}-\hat{j}+\hat{k})$ $(1)$
and $\vec{r}=2 \hat{i}+\hat{j}-\hat{k}+\mu(3 \hat{i}-5 \hat{j}+2 \hat{k})$ $(2)$