The distance of the point $B(1, 2, 3)$ from the line passing through $A(4, 2, 2)$ and parallel to the vector $\vec{c} = 2i + 3j + 6k$ is

Let $P(\alpha, \beta, \gamma)$ be the point on the line $\frac{x-1}{2} = \frac{y+1}{-3} = \frac{z}{1}$ at a distance $4\sqrt{14}$ from the point $(1, -1, 0)$ and nearer to the origin. Then the shortest distance between the lines $\frac{x-\alpha}{1} = \frac{y-\beta}{2} = \frac{z-\gamma}{3}$ and $\frac{x+5}{2} = \frac{y-10}{1} = \frac{z-3}{1}$ is equal to

The line $L_1$ is parallel to the vector $\vec{a} = -3 \hat{i} + 2 \hat{j} + 4 \hat{k}$ and passes through the point $(7, 6, 2)$,and the line $L_2$ is parallel to the vector $\vec{b} = 2 \hat{i} + \hat{j} + 3 \hat{k}$ and passes through the point $(5, 3, 4)$. The shortest distance between the lines $L_1$ and $L_2$ is:

Find the cartesian equation of the line which passes through the point $(-2, 4, -5)$ and is parallel to the line given by $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}$.

The angle between a line with direction ratios $2, 2, 1$ and the line joining the points $(3, 1, 4)$ and $(7, 2, 12)$ is

The Cartesian equation of a line is $3x + 1 = 6y - 2 = -z + 1$. Find its vector equation.