The distance of the point $P(4, 6, -2)$ from the line passing through the point $(-3, 2, 3)$ and parallel to a line with direction ratios $3, 3, -1$ is equal to:

The angle between the lines $3x + 2y + z = 0 = x + y - 2z$ and $2x - y - z = 0 = 7x + 10y - 8z$ is

The equation of the line passing through the points $(3, 4, -7)$ and $(6, -1, 1)$ is

The equation of the line,passing through $A(1, 2, 3)$ and perpendicular to the vectors $2 \hat{i} + \hat{j} - \hat{k}$ and $\hat{i} + 3 \hat{j} + 2 \hat{k}$,is

Let the shortest distance between the lines $\frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4}$ be $3 \sqrt{30}$. Then the positive value of $5 \alpha+\beta$ is

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