The line $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{-1}$ intersects the curve $xy = c^2, z = 0$ if $c$ is equal to

$A$ line $l$ passes through the origin and is perpendicular to the lines $l_1 = (3 + t)\hat{i} + (-1 + 2t)\hat{j} + (4 + 2t)\hat{k}$ and $l_2 = (3 + 2s)\hat{i} + (3 + 2s)\hat{j} + (2 + s)\hat{k}$.
Statement $1$: Line $l$ and $l_2$ are coplanar lines.
Statement $2$: Line $l$ and $l_2$ are intersecting lines.

The distance of the point $(1, 0, 2)$ from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ is

If the line $r = a + t b$ is parallel to the plane $r = c + l d + m e$, then

If the lines $\overrightarrow{r} = (\hat{i} - \hat{j} + \hat{k}) + \lambda(3\hat{j} - \hat{k})$ and $\overrightarrow{r} = (\alpha\hat{i} - \hat{j}) + \mu(2\hat{i} - 3\hat{k})$ are coplanar,then the distance of the plane containing these two lines from the point $(\alpha, 0, 0)$ is

$A$ line drawn from the point $A(1, 3, 2)$ parallel to the line $\frac{x}{2} = \frac{y}{4} = \frac{z}{1}$ intersects the plane $3x + y + 2z = 5$ at point $B$. Then the coordinates of point $B$ are: