Question

If the lines $\frac{x}{1}=\frac{y-1}{-1}=\frac{z+2}{2},\frac{x}{2}=\frac{y-1}{1}=\frac{z+2}{1},\frac{x}{-1}=\frac{y-1}{-1}=\frac{z+2}{1}$ intersect the plane $x+y-z=0$ at $P,Q,R$ respectively, then the area (in sq. units) of triangle $PQR$ is

• A. $\sqrt{3}$
• B. $\frac{4}{7}\sqrt{3}$
• C. $\frac{9}{8}\sqrt{3}$
• D. $\frac{9}{4}\sqrt{3}$

Answer 1

Answer: (C)

Coordinates of $P,Q,R$ are $\left(r_1,-r_1+1,2r_1-2\right),$ $\left(2r_2,r_2+1,r_2-2\right),$ $\left(-r_3,-r_3+1,r_3-2\right)$ respectively.
It must satisfy the plane, hence $r_1=\frac{3}{2},r_2=\frac{-3}{2},r_3=+1.$ So points $P,Q,R$ are $\left(\frac{3}{2},-\frac{1}{2},1\right),\left(-3,-\frac{1}{2},\frac{-7}{2}\right)$ $\left(-1,0,-1\right)$ respectively.
$\overrightarrow{\mathrm{PQ}}=-\frac{9}{2}\hat{i}-\frac{9}{2}\hat{k}$
$\overrightarrow{\mathrm{PR}}=\frac{-5}{2}\hat{ı}+\frac{1}{2}\hat{j}-2\hat{k}$
Area $=\frac{1}{2}\left|\overrightarrow{\mathrm{PQ}}\times \overrightarrow{\mathrm{PR}}\right|$
$=\left|\frac{1}{8}\right|\left|\begin{array}{ccc}i& \hat{j}& \hat{k}\\ -9& 0& -9\\ -5& 1& -4\end{array}\right|$
$=\frac{1}{8}\left|\begin{array}{ccc}9\hat{i}& +9\hat{j}& -9\hat{k}\end{array}\right|=\frac{9\sqrt{3}}{8}$ sq. units