JEE Math Practice Question
The vertices B and C of a $$\Delta $$ABC lie on the line, $${{x + 2} \over 3} = {{y - 1} \over 0} = {z \over 4}$$ such that BC = 5 units. Then the area (in sq. units) of this triangle, given that the point A(1, –1, 2), is :
- A.6
- B.$$5\sqrt {17} $$
- C.$$\sqrt {34} $$
- D.$$2\sqrt {34} $$
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