JEE Math Practice Question
Let the foot of the perpendicular from the point (1, 2, 4) on the line $${{x + 2} \over 4} = {{y - 1} \over 2} = {{z + 1} \over 3}$$ be P. Then the distance of P from the plane $$3x + 4y + 12z + 23 = 0$$ is :
- A.5
- B.$${{50} \over {13}}$$
- C.4
- D.$${{63} \over {13}}$$
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