The vertices B and C of a $\Delta ABC$  lie on 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)

Given B & C of $\Delta ABC$ lie on the line $\frac{x+2}{3}=\frac{y-1}{0}=\frac{z}{y}BC=5$

A(1,-1,2)

Area of triangle $ABC=\frac{1}{2}\times \left(BC\right)\times$ (ter distance from a to line)

let us take a point on the line

$P\left(3\lambda -2,1,4\lambda \right)$

AP line is perpendicular to direction of given line

$\overrightarrow{AP}=\left(3\lambda -3,2,4\lambda -2\right)i.e\left(3\lambda -3\hat{i}+2\hat{j}+\left(4\lambda \right)\right)\hat{k}$

$\overrightarrow{AP}.\overrightarrow{L}=0$

$\left(3\lambda -3\right)\hat{i}+2\hat{j}+\left(4\lambda -2\right)\hat{k}.\left(3\hat{i}+0\hat{j}+4\hat{k}\right)=0$

$9\lambda -9+10\lambda -8=0$

$25\lambda =17$

$\lambda =\frac{17}{25}$

$P\left(\frac{1}{25},1,\frac{68}{25}\right)$

$\left|\overline{AP}\right|=\sqrt{{\left(\frac{24}{25}\right)}^2+2^2+{\left(\frac{18}{25}\right)}^2}$

Area of triangle ABC $=\frac{1}{x}\times 5\times \left|\overline{AD}\right|$ $=\sqrt{34}$