The equation of the line passing through $(-4,3,1)$ parallel to the plane $x+2 y-z-5=0$ and intersecting the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z-2}{-1}$ is

If the plane $3x + y + 2z + 6 = 0$ is parallel to the line ${{3x - 1} \over {2b}} = 3 - y = {{z - 1} \over a}$, then the value of $3a + 3b$ is

Angle between the diagonals of a cube is

Consider the two lines

${L_1}:{{x + 1} \over 3} = {{y + 2} \over 1} = {{z + 1} \over 2}$ and ${L_2}:{{x - 2} \over 1} = {{y + 2} \over 2} = {{z - 3} \over 3}$

The unit vector perpendicular to both the lines L₁ and L₂ is

The distance between the line $r = 2\widehat i - 2\widehat j + 3\widehat k + \lambda (\widehat i - \widehat j + 4\widehat k)$ and the plane $a\,.\,(\widehat i + 5\widehat j + \widehat k) = 5$ is

A line passing through P(3, 7, 1) and R(2, 5, 7) meet the plane 3x + 2y + 11z $-$ 9 = 0 at Q. Then PQ is equal to