The straight line $\frac{x-3}{3}=\frac{y-2}{1}=\frac{z-1}{0}$ is

The plane $$2 x-y+3 z+5=0$$ is rotated through $$90^{\circ}$$ about its line of intersection with the plane $$x+y+z=1$$. The equation of the plane in new position is

If the relation between the direction ratios of two lines in $$\mathbb{R}^3$$ are given by

$$l+\mathrm{m}+\mathrm{n}=0,2 l \mathrm{~m}+2 \mathrm{mn}-l \mathrm{n}=0$$

then the angle between the lines is ($$l, \mathrm{~m}, \mathrm{n}$$ have their usual meaning)

Angle between two diagonals of a cube will be

If the distance between the plane $$\alpha x - 2y + z = k$$ and the plane containing the lines $${{x - 1} \over 2} = {{y - 2} \over 3} = {{z - 3} \over 4}$$ and $${{x - 2} \over 3} = {{y - 3} \over 4} = {{z - 4} \over 5}$$ is $$\sqrt 6 $$, then $$|k|$$ is

The angle between a normal to the plane $$2x - y + 2z - 1 = 0$$ and the X-axis is

The equation of the plane through the intersection of the planes x + y + z = 1 and 2x + 3y $$-$$ z + 4 = 0 and parallel to the x-axis is

The line $$x - 2y + 4z + 4 = 0$$, $$x + y + z - 8 = 0$$ intersect the plane $$x - y + 2z + 1 = 0$$ at the point