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Question Id : 15807 | Context : JEE Main 2019 (9 April Evening Shift)
The vertices $B$ and $C$ of a $\triangle ABC$ lie on the line $\dfrac{x+2}{3}=\dfrac{y-1}{0}=\dfrac{z}{4}$ such that $BC=5$ units. Then the area (in sq. units) of this triangle, given that the point $A(1,-1,2)$, is:

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