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