The equation of the perpendicular from the point $(\alpha, \beta, \gamma)$ to the plane $ax + by + cz + d = 0$ is

The acute angle $\theta$ between the line $\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} + \hat{j} + \hat{k})$ and the plane $\vec{r} \cdot (2\hat{i} + p\hat{j} + \hat{k}) = 8$ is given by $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$, where $\vec{b} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{n} = 2\hat{i} + p\hat{j} + \hat{k}$. If $\theta = \sin^{-1} \left( \frac{\sqrt{2}}{3} \right)$, then the value$(s)$ of $p$ is/are:

At what point does the line joining the points $(2, -3, 1)$ and $(3, -4, -5)$ intersect the plane $2x + y + z = 7$?

The point$(s)$ on the line $\vec r = \hat i + \hat j + \hat k + t(\hat i + 3\hat j - \hat k)$ at a distance of $3 \ units$ from the plane $\vec r \cdot (\hat i + 2\hat j + 2\hat k) + 2 = 0$ are

The angle between the line $r = (i + 2j - k) + \lambda (i - j + k)$ and the normal to the plane $r \cdot (2i - j + k) = 4$ is

Let a plane $P$ contain two lines $\overrightarrow{r} = \hat{i} + \lambda(\hat{i} + \hat{j}), \lambda \in R$ and $\overrightarrow{r} = -\hat{j} + \mu(\hat{j} - \hat{k}), \mu \in R$. If $Q(\alpha, \beta, \gamma)$ is the foot of the perpendicular drawn from the point $M(1, 0, 1)$ to $P$,then $3(\alpha + \beta + \gamma)$ equals