The vertices B and C of a $\Delta ABC$ 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 $A\left(1,-1,2\right)$, is

The given line is $\frac{x+2}{3}=\frac{y-1}{0}=\frac{z}{4}=\lambda \left(say\right)$. So, the coordinates of any point D on the given line are  $\left(3\lambda -2,1,4\lambda \right)$. Let AD is perpendicular to the given line.

$\therefore \overline{AD}.\left(3\widehat{i}+4\widehat{k}\right)=0$

$\Rightarrow \left(3\lambda -2-1\right)\left(3\right)+\left(1+1\right)\left(0\right)+\left(4\lambda -2\right)4=0$
$\Rightarrow 9\lambda -9+16\lambda -8=0\Rightarrow 25\lambda =17\Rightarrow \lambda =\frac{17}{25}$

$\therefore -\frac{24}{25}\widehat{i}+2\widehat{j}+\frac{18}{25}\widehat{k}$

So, $\left|\overline{AD}\right|=\sqrt{{\left(\frac{-24}{25}\right)}^2+{\left(2\right)}^2+{\left(\frac{18}{25}\right)}^2}=\sqrt{\frac{576}{625}+4+\frac{324}{625}}=\sqrt{\frac{900+2500}{625}}$

$=\sqrt{\frac{3400}{625}}=\frac{\sqrt{34}}{25}\times 10=\frac{2}{5}\sqrt{34}$

Now, area of $\Delta ABC=\frac{1}{2}\times BC\times AD=\frac{1}{2}\times 5\times \frac{2}{5}\sqrt{34}=\sqrt{34}sq.units$