The distance of the point $(7,5,2)$ from the plane $3x+4y+z-8=0$ measured parallel to the line $\frac{x-1}{3}=\frac{y-2}{6}=\frac{z+1}{2}$ is:

If the distance of the point $P(1, -2, 1)$ from the plane $x + 2y - 2z = \alpha$, where $\alpha > 0$, is $5$, then the foot of the perpendicular from $P$ to the plane is

Let $P_{1}: \vec{r} \cdot(2 \hat{i} + \hat{j} - 3 \hat{k}) = 4$ be a plane. Let $P_{2}$ be another plane which passes through the points $(2, -3, 2)$,$(2, -2, -3)$,and $(1, -4, 2)$. If the direction ratios of the line of intersection of $P_{1}$ and $P_{2}$ are $16, \alpha, \beta$,then the value of $\alpha + \beta$ is equal to

The line passing through the points $(1, 1, -1)$ and $(3, -1, 0)$ makes an angle of $\operatorname{Tan}^{-1}\left(\frac{1}{\sqrt{8}}\right)$ with the plane $\sqrt{\lambda} x + 3y + 6z = 17$. Then $\lambda =$

The coordinates of the point where the line $\frac{x - 6}{-1} = \frac{y + 1}{0} = \frac{z + 3}{4}$ meets the plane $x + y - z = 3$ are

Consider the line $L$ given by the equation $\frac{x-3}{2}=\frac{y-1}{1}=\frac{z-2}{1}$. Let $Q$ be the mirror image of the point $P_0(2,3,-1)$ with respect to $L$. Let a plane $P$ be such that it passes through $Q$,and the line $L$ is perpendicular to $P$. Then which of the following points is on the plane $P$?