If the two lines $${l_1}:{{x - 2} \over 3} = {{y + 1} \over {-2}},\,z = 2$$ and $${l_2}:{{x - 1} \over 1} = {{2y + 3} \over \alpha } = {{z + 5} \over 2}$$ are perpendicular, then an angle between the lines l₂ and $${l_3}:{{1 - x} \over 3} = {{2y - 1} \over { - 4}} = {z \over 4}$$ is :

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 mirror image of the point $$\left( {2, - {1 \over 2},2} \right)$$ in the rotated plane is B(a, b, c), then :

Let p be the plane passing through the intersection of the planes $$\overrightarrow r \,.\,\left( {\widehat i + 3\widehat j - \widehat k} \right) = 5$$ and $$\overrightarrow r \,.\,\left( {2\widehat i - \widehat j + \widehat k} \right) = 3$$, and the point (2, 1, $$-$$2). Let the position vectors of the points X and Y be $$\widehat i - 2\widehat j + 4\widehat k$$ and $$5\widehat i - \widehat j + 2\widehat k$$ respectively. Then the points :

Let Q be the mirror image of the point P(1, 0, 1) with respect to the plane S : x + y + z = 5. If a line L passing through (1, $$-$$1, $$-$$1), parallel to the line PQ meets the plane S at R, then QR² is equal to :