For real numbers $\alpha$ and $\beta \neq 0$,if the point of intersection of the straight lines $\frac{x-\alpha}{1}=\frac{y-1}{2}=\frac{z-1}{3}$ and $\frac{x-4}{\beta}=\frac{y-6}{3}=\frac{z-7}{3}$ lies on the plane $x+2y-z=8$,then $\alpha-\beta$ is equal to:

$A$ ray of light passing through the point $A(1, 2, 3)$ strikes the plane $x+y+z=12$ at $B$ and on reflection passes through $C(3, 5, 9)$,then $OB$ is equal to

Let a line $L$ passing through the point $P(1, 1, 1)$ be perpendicular to the lines $\frac{x-4}{4}=\frac{y-1}{1}=\frac{z-1}{1}$ and $\frac{x-17}{1}=\frac{y-71}{1}=\frac{z}{0}$. Let the line $L$ intersect the $yz$-plane at the point $Q$. Another line parallel to $L$ and passing through the point $S(1, 0, -1)$ intersects the $yz$-plane at the point $R$. Then the square of the area of the parallelogram $PQRS$ is equal to . . . . . . .

If the equation of the plane passing through the line of intersection of the planes $2x - 7y + 4z - 3 = 0$ and $3x - 5y + 4z + 11 = 0$ and the point $(-2, 1, 3)$ is $ax + by + cz - 7 = 0$,then the value of $2a + b + c - 7$ is

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

The equation of the plane passing through the line $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{4}$ and perpendicular to the plane $x+2y+z=12$ is given by $ax+by+cz+4=0$. Then: