The lines $\frac{x - a + d}{\alpha - \delta} = \frac{y - a}{\alpha} = \frac{z - a - d}{\alpha + \delta}$ and $\frac{x - b + c}{\beta - \gamma} = \frac{y - b}{\beta} = \frac{z - b - c}{\beta + \gamma}$ are coplanar. Find the equation of the plane in which they lie.

If the equation of the plane passing through the point $(2, -1, 3)$ and perpendicular to the planes $3x - 2y + z = 9$ and $x + y + z = 9$ is $x + by + cz + d = 0$,then $d =$

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

What is the distance between the line $\frac{x - 1}{3} = \frac{y + 2}{-2} = \frac{z - 1}{2}$ and the plane $2x + 2y - z = 6$?

If $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-7}{2}$ lies in the plane $ax+by+z=7$,then $a+b=$

Let the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lie in the plane $x+3y-\alpha z+\beta=0$. Then the value of $(\beta-\alpha)$ is equal to