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 $k = . . . . .$.

The distance of the point $-\hat{i} + 2\hat{j} + 6\hat{k}$ from the straight line that passes through the point $2\hat{i} + 3\hat{j} - 4\hat{k}$ and is parallel to the vector $6\hat{i} + 3\hat{j} - 4\hat{k}$ is

If $(a, b, c)$ are the direction ratios of a line joining the points $(4, 3, -5)$ and $(-2, 1, -8)$,then the point $P(a, 3b, 2c)$ lies on the plane:

The length of the perpendicular from the point $A(1, -2, -3)$ on the line $\frac{x-1}{2} = \frac{y+3}{-1} = \frac{z+1}{-2}$ is (in $\text{ units}$)

Let the line $L$ intersect the lines $x-2=-y=z-1$ and $2(x+1)=2(y-1)=z+1$,and be parallel to the line $\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}$. Then which of the following points lies on $L$?

The point of intersection of the lines $l_1: r(t) = (i - 6j + 2k) + t(i + 2j + k)$ and $l_2: R(u) = (4j + k) + u(2i + j + 2k)$ is