The lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{-k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar,if

The equation of the planes passing through the line of intersection of the planes $3x - y - 4z = 0$ and $x + 3y + 6 = 0$ whose distance from the origin is $1$,are

The direction ratios of the normal to the plane passing through the points $(1, 2, -3)$, $(-1, -2, 1)$ and parallel to $\frac{x-2}{2}=\frac{y+1}{3}=\frac{z}{4}$ are:

The distance of the point $(1, 3, -7)$ from the plane passing through the point $(1, -1, -1)$ and having a normal perpendicular to both the lines $\frac{x - 1}{1} = \frac{y + 2}{-2} = \frac{z - 4}{3}$ and $\frac{x - 2}{2} = \frac{y + 1}{-1} = \frac{z + 7}{-1}$ is . . . .

Consider the planes $3x - 6y - 2z = 15$ and $2x + y - 2z = 5$.
$STATEMENT-1$ : The parametric equations of the line of intersection of the given planes are $x = 3 + 14t, y = 1 + 2t, z = 15t$ because
$STATEMENT-2$ : The vector $14\hat{i} + 2\hat{j} + 15\hat{k}$ is parallel to the line of intersection of the given planes.

If the line segment joining the points $P(2, 4, 1)$ and $Q(3, 8, 1)$ is divided by the plane $3x - ky - 6z = 0$ externally in the ratio $4:5$, then $k=$