The square of the distance of the point of intersection of the lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(\hat{i} - \hat{j})$ and $\vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + \hat{k})$ from the origin is:

The shortest distance between the lines $x+1=2y=-12z$ and $x=y+2=6z-6$ is

The perpendicular distance from the point $P(-1, 1, 0)$ to the line joining the points $A(0, 2, 4)$ and $B(3, 0, 1)$ is

The length of the perpendicular drawn from the point $2 \hat{i} - \hat{j} + 5 \hat{k}$ to the line $\vec{r} = (11 \hat{i} - 2 \hat{j} - 8 \hat{k}) + \lambda(10 \hat{i} - 4 \hat{j} - 11 \hat{k})$ is

Consider the line $L$ passing through the points $(1, 2, 3)$ and $(2, 3, 5)$. The distance of the point $A\left(\frac{11}{3}, \frac{11}{3}, \frac{19}{3}\right)$ from the line $L$ along the line $\frac{3x-11}{2} = \frac{3y-11}{1} = \frac{3z-19}{2}$ is equal to:

Let the point,on the line passing through the points $P(1, -2, 3)$ and $Q(5, -4, 7)$,farther from the origin and at a distance of $9$ units from the point $P$,be $(\alpha, \beta, \gamma)$. Then $\alpha^2 + \beta^2 + \gamma^2$ is equal to: