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:

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

The equation of the line passing through the point $(2, 3, 1)$ and parallel to the line of intersection of the planes $x - 2y - z + 5 = 0$ and $x + y + 3z = 6$ is:

The centre of the circle given by $r \cdot (i + 2j + 2k) = 15$ and $|r - (j + 2k)| = 4$ is

The direction cosines of the line of intersection of the planes $x+2y+z-4=0$ and $2x-y+z-3=0$ are

$L$ is a line passing through the point $A(1, 0, -3)$ and parallel to a line having direction ratios $0, 1, -2$. $P$ is a point on the line $L$ which is at a minimum distance from the plane $2x + 3y + 5z = 1$. Then,the equation of the plane through $P$ and perpendicular to $AP$ is