JEE Math Practice Question
The distance of the point $$(1, 0, 2)$$ from the point of intersection of the line $${{x - 2} \over 3} = {{y + 1} \over 4} = {{z - 2} \over {12}}$$ and the plane $$x - y + z = 16,$$ is :
- A.$$3\sqrt {21} $$
- B.$$13$$
- C.$$2\sqrt {14} $$
- D.$$8$$
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