JEE Math Practice Question
If the line $${{x - 2} \over 3} = {{y + 1} \over 2} = {{z - 1} \over { - 1}}$$ intersects the plane 2x + 3y – z + 13 = 0 at a point P and the plane 3x + y + 4z = 16 at a point Q, then PQ is equal to :
- A.$$2\sqrt 7 $$
- B.14
- C.$$2\sqrt {14} $$
- D.$$\sqrt {14} $$
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