For real numbers $\alpha$ and $\beta$ $\ne$ 0, if the point of intersection of the straight lines

${{x - \alpha } \over 1} = {{y - 1} \over 2} = {{z - 1} \over 3}$ and ${{x - 4} \over \beta } = {{y - 6} \over 3} = {{z - 7} \over 3}$, lies on the plane x + 2y $-$ z = 8, then $\alpha$ $-$ $\beta$ is equal to :

Let the line $l: x=\frac{1-y}{-2}=\frac{z-3}{\lambda}, \lambda \in \mathbb{R}$ meet the plane $P: x+2 y+3 z=4$ at the point $(\alpha,…

Let the line $\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$ intersect the lines $\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$ and…

Let the plane 2x + 3y + z + 20 = 0 be rotated through a right angle about its line of intersection with the plane x $-$ 3y + 5z = 8. If the…

The acute angle between the planes P1 and P2, when P1 and P2 are the planes passing through the intersection of the planes $5x + 8y + 13z -…

The distance of the point (7, $-$3, $-$4) from the plane passing through the points (2, $-$3, 1), ($-$1, 1, $-$2) and (3, $-$4, 2) is :