Find the equation of the plane passing through the intersection of the planes $P_1$ and $P_2$ and parallel to the line $L$,where:
$P_1 : 3x + 2y + 5z + 1 = 0$
$P_2 : x + y + z + 2 = 0$
$L : \frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$

The point of intersection of the line $x+1=\frac{y+3}{3}=\frac{-z+2}{2}$ with the plane $3x+4y+5z=10$ 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

The position vectors of the points $A$ and $B$ are respectively $\hat{i}+2 \hat{j}$ and $2 \hat{i}+\hat{j}+\hat{k}$. If the points $P$ and $Q$ are respectively the orthogonal projections of $A$ and $B$ on the plane $x+y+z=3$, then $P Q=$

The angle between the line $\frac{x - 1}{-2} = \frac{y - 2}{1} = \frac{z + 1}{2}$ and the plane $3x + 2y + 6z = 1$ is:

Consider a pyramid $OPQRS$ located in the first octant $(x \geq 0, y \geq 0, z \geq 0)$ with $O$ as the origin,and $OP$ and $OR$ along the $x$-axis and the $y$-axis,respectively. The base $OPQR$ of the pyramid is a square with $OP=3$. The point $S$ is directly above the mid-point $T$ of diagonal $OQ$ such that $TS=3$. Then:
$(A)$ the acute angle between $OQ$ and $OS$ is $\frac{\pi}{3}$
$(B)$ the equation of the plane containing the triangle $OQS$ is $x-y=0$
$(C)$ the length of the perpendicular from $P$ to the plane containing the triangle $OQS$ is $\frac{3}{\sqrt{2}}$
$(D)$ the perpendicular distance from $O$ to the straight line containing $RS$ is $\sqrt{\frac{15}{2}}$