The equation of the line that passes through the origin and is parallel to the $X$-axis is . . . . . . .

The distance of the point $P(-2, 4, -5)$ from the line $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}$ is

The point of intersection of the lines represented by $r=(\hat{i}+2 \hat{j}-\hat{k})+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})$ and $r=(-\hat{i}-3 \hat{j}+7 \hat{k})+\mu(\hat{i}+2 \hat{j}-\hat{k})$ is

Find the length of the perpendicular drawn from the point $(1, 2, 3)$ to the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{-2}$.

$\triangle ABC$ is formed by $A(1, 8, 4)$,$B(0, -11, 4)$ and $C(2, -3, 1)$. If $D$ is the foot of the perpendicular from $A$ to $BC$,then the coordinates of $D$ are

The shortest distance between the lines $\vec{r} = (\frac{1}{3}\hat{i} + 2\hat{j} + \frac{8}{3}\hat{k}) + \lambda(2\hat{i} - 5\hat{j} + 6\hat{k})$ and $\vec{r} = (-\frac{2}{3}\hat{i} - \frac{1}{3}\hat{k}) + \mu(\hat{i} - \hat{k})$, where $\lambda, \mu \in R$, is: