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