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

In what ratio does the plane $\vec{r} \cdot (\hat{i} - 2\hat{j} + 3\hat{k}) = 17$ divide the line segment joining the points $-2\hat{i} + 4\hat{j} + 7\hat{k}$ and $3\hat{i} - 5\hat{j} + 8\hat{k}$?

The equation of the locus of a point $P(x, y, z)$ such that its distance from the $X$-axis is equal to its distance from the plane $x+z=1$ is

Let $Q$ be the foot of the perpendicular from the point $P(7, -2, 13)$ on the plane containing the lines $\frac{x+1}{6} = \frac{y-1}{7} = \frac{z-3}{8}$ and $\frac{x-1}{3} = \frac{y-2}{5} = \frac{z-3}{7}$. Then $(PQ)^{2}$ is equal to ..... .

Find the distance of the point $-\hat{i} - 5\hat{j} - 10\hat{k}$ from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 5$.

Let the line $\ell: x = \frac{1-y}{-2} = \frac{z-3}{\lambda}, \lambda \in R$ meet the plane $P: x + 2y + 3z = 4$ at the point $(\alpha, \beta, \gamma)$. If the angle between the line $\ell$ and the plane $P$ is $\cos^{-1}\left(\sqrt{\frac{5}{14}}\right)$,then $\alpha + 2\beta + 6\gamma$ is equal to