The plane passing through the points (1, 2, 1),
(2, 1, 2) and parallel to the line, 2x = 3y, z = 1
also passes through the point :

The mirror image of the point (1, 2, 3) in a plane is

$$\left( { - {7 \over 3}, - {4 \over 3}, - {1 \over 3}} \right)$$. Which of the following points lies on this plane ?

The shortest distance between the lines

$${{x - 3} \over 3} = {{y - 8} \over { - 1}} = {{z - 3} \over 1}$$ and

$${{x + 3} \over { - 3}} = {{y + 7} \over 2} = {{z - 6} \over 4}$$ is :

Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be any point (2, 1, 6). Then the image of R in the plane P is :