The distance of the point $(2, 3, 4)$ from the plane $3x - 6y + 2z + 11 = 0$ is

If $4x + 4y - kz = 0$ is the equation of the plane through the origin that contains the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4},$ then $k =$

The distance of the point $(1, -5, 9)$ from the plane $x - y + z = 5$ measured along the line $x = y = z$ is . . . . . . units.

Statement-$1$: The point $A(1, 0, 7)$ is the reflection of the point $B(1, 6, 3)$ in the line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$.
Statement-$2$: The line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$ is the perpendicular bisector of the line segment joining $A(1, 0, 7)$ and $B(1, 6, 3)$.

Find the distance between the line $\vec{r} = (2\hat{i} - 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} + 4\hat{k})$ and the plane $\vec{r} \cdot (\hat{i} + 5\hat{j} + \hat{k}) = 5$.

The equation of the plane passing through the point $(1,1,1)$ and through the line of intersection of $x+2y-z+1=0$ and $3x-y-4z+3=0$ is