The equation of the plane,passing through the intersection of the planes $x+y+z=1$ and $2x+3y-z+4=0$ and parallel to $Y$-axis 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 $\sqrt{\frac{15}{2}}$

Find the vector equation of the line passing through $(1, 2, 3)$ and parallel to the planes $\vec{r} \cdot (\hat{i} - \hat{j} + 2\hat{k}) = 5$ and $\vec{r} \cdot (3\hat{i} + \hat{j} + \hat{k}) = 6$.

$A$ plane $ax+by+cz+1=0$ is perpendicular to the two planes $2x-2y+z=0$ and $x-y+2z=4$ and passes through the point $(1, -2, 1)$. Then $a+b-c=$

The value of $k$,such that the line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$ lies on the plane $2x-4y+z=7$,is

Let $A$ be a point having position vector $\bar{i}-3 \bar{j}$ and $\bar{r}=(\bar{i}-3 \bar{j})+t(\bar{j}-2 \bar{k})$ be a line. If $P$ is a point on this line and is at a minimum distance from the plane $\bar{r} \cdot(2 \bar{i}+3 \bar{j}+5 \bar{k})=0$, then the equation of the plane through $P$ and perpendicular to $AP$ is: