Perpendiculars are drawn from points on the line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}$ to the plane $x+y+$ $z=3$. The foot of perpendiculars lie on the line

The equation of a plane passing through the line of intersection of the planes $$x+2y+3z=2$$ and $$x-y+z=3$$ and at a distance $${2 \over {\sqrt 3 }}$$ from the point $$(3, 1, -1)$$ is

The point $$P$$ is the intersection of the straight line joining the points $$Q(2, 3, 5)$$ and $$R(1, -1, 4)$$ with the plane $$5x-4y-z=1.$$ If $$S$$ is the foot of the perpendicular drawn from the point $$T(2, 1, 4)$$ to $$QR,$$ then the length of the line segment $$PS$$ is

Equation of the plane containing the straight line $${x \over 2} = {y \over 3} = {z \over 4}$$ and perpendicular to the plane containing the straight lines $${x \over 3} = {y \over 4} = {z \over 2}$$ and $${x \over 4} = {y \over 2} = {z \over 3}$$ is