If $4x + 4y - kz = 0$ is the equation of the plane through the origin that contains the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4},$ then $k =$

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 $\sqrt{\frac{15}{2}}$

If the plane passing through the points $\hat{i}+\hat{j}+\hat{k}$, $2\hat{i}-\hat{k}$ and the origin meets the line passing through the points $\hat{i}+3\hat{j}-2\hat{k}$ and $\hat{i}-\hat{j}+3\hat{k}$ at the point $A$, then $A=$

The equation of the plane passing through the intersection of the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-5}{-3}$ and $\frac{x+5}{3}=\frac{y-4}{-1}=\frac{z+3}{4}$ and parallel to the $xy$-plane is

If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect,then $k$ is equal to:

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