For a positive real number $p$,if the perpendicular distance from a point $-\hat{i} + p\hat{j} - 3\hat{k}$ to the plane $\vec{r} \cdot (2\hat{i} - 3\hat{j} + 6\hat{k}) = 7$ is $6$ units,then $p=$

The equation of the line passing through the intersection of the plane $x+2y+3z=4$ and the line $\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-1}{-1}$ and parallel to the vector $(2\hat{i}-3\hat{j}) \times (\hat{i}+2\hat{j}-\hat{k})$ is

The plane containing the line $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and parallel to the line $\frac{x}{1} = \frac{y}{1} = \frac{z}{4}$ passes through the point

The line $\frac{x + 3}{3} = \frac{y - 2}{-2} = \frac{z + 1}{1}$ and the plane $4x + 5y + 3z - 5 = 0$ intersect at a point

Find the coordinates of the foot of the perpendicular drawn from the point $P(-1, 1, 2)$ to the plane $2x - 3y + z - 11 = 0$.

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