If $\frac{x - 1}{l} = \frac{y - 2}{m} = \frac{z + 1}{n}$ is the equation of the line passing through the points $(1, 2, -1)$ and $(-1, 0, 1)$,then the direction ratios $(l, m, n)$ are:

Find the coordinates of the foot of the perpendicular drawn from the point $A(1, 0, 3)$ to the line joining the points $B(4, 7, 1)$ and $C(3, 5, 3)$.

The shortest distance between the skew lines $\vec{r}=(2 \hat{i}-\hat{j})+t(\hat{i}+2 \hat{k})$ and $\vec{r}=(-2 \hat{i}+\hat{k})+s(\hat{i}-\hat{j}-\hat{k})$ is

If the lines $\frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2}$ and $\frac{x - 1}{3k} = \frac{y - 5}{1} = \frac{z - 6}{-5}$ are perpendicular to each other,then $k = \dots$

The position vector of the point of intersection of the line joining the points $2 \hat{i}-\hat{j}+6 \hat{k}$ and $3 \hat{i}-\hat{j}-7 \hat{k}$, and the line joining the points $2 \hat{i}+\hat{j}-6 \hat{k}$ and $3 \hat{i}-\hat{j}-7 \hat{k}$ is:

Let $P$ be the image of the point $Q(7,-2,5)$ in the line $L: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ and $R(5, p, q)$ be a point on $L$. Then the square of the area of $\triangle P Q R$ is $\qquad$