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