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 $\left(r_{1} , - r_{1} + 1 , 2 r_{1} - 2\right),$ $\left(2 r_{2} , r_{2} + 1 , r_{2} - 2\right),$ $\left(- r_{3} , - r_{3} + 1 , r_{3} - 2\right)$ 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.
$\overset{ \rightarrow }{P Q}=-\frac{9}{2}\hat{i}-\frac{9}{2}\hat{k}$
$\overset{ \rightarrow }{P R}=\frac{- 5}{2}\hat{ı}+\frac{1}{2}\hat{j}-2\hat{k}$
Area $=\frac{1}{2}\left|\overset{ \rightarrow }{P Q} \times \overset{ \rightarrow }{P R}\right|$
$=\left|\frac{1}{8}\right|\begin{vmatrix} i & \hat{j} & \hat{k} \\ -9 & 0 & -9 \\ -5 & 1 & -4 \end{vmatrix}$
$=\frac{1}{8}\begin{vmatrix} 9\hat{i} & +9\hat{j} & -9\hat{k} \end{vmatrix}=\frac{9 \sqrt{3}}{8}$ sq. units