Question

If two distinct point $Q,R$ lie on the line of intersection of the planes $-x+2y-z=0$ and $3x-5y+2z=0$ and $PQ=PR=\sqrt{18}$ where the point $P$ is $(1,-2,3)$, then the area of the triangle $PQR$ is equal to

• A. $\frac{2}{3}\sqrt{38}$
• B. $\frac{4}{3}\sqrt{38}$
• C. $\frac{8}{3}\sqrt{38}$
• D. $\sqrt{\frac{152}{3}}$

Answer 1

Answer: (B)

$-x+2y-z=0$
$3x-5y+2z=0$
$\vec{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1& 2& -1\\ 3& -5& 2\end{array}\right|$
$=\hat{i}(-1)-\hat{j}(1)+\hat{k}(-1)$
$\vec{n}=-\hat{i}-\hat{j}-\hat{k}$
Equation of LOI is $\frac{x}{1}=\frac{y}{1}=\frac{z}{1}$
DR: of PT $\to \alpha -1,\alpha +2,\alpha -3$
$DR:$ of $QR\to 1,1,1$
$\Rightarrow (\alpha -1)\times 1+(\alpha +2)\times 1+(\alpha -3)\times 1=0$
$3\alpha =2$
$\alpha =\frac{2}{3}$
$P{T}^2=\frac{1}{9}+\frac{64}{9}+\frac{49}{9}$
$P{T}^2=\frac{114}{9}$
$PT=\frac{\sqrt{114}}{3}$
$\cos \theta =\frac{\sqrt{114}}{3}\times \frac{1}{3\sqrt{2}}=\frac{\sqrt{57}}{9}=\frac{\sqrt{19\times 3}}{3\times 3}$
$=\frac{\sqrt{19}}{3\sqrt{3}}$
$\cos 2\theta =\frac{2\times 19}{27}-1=\frac{11}{27}$
$\sin 2\theta =\sqrt{1-{\left(\frac{11}{27}\right)}^2}=\frac{\sqrt{38}\sqrt{16}}{27}$
$=\frac{4}{27}\sqrt{38}$
Area $=\frac{1}{2}\times \sqrt{18}\sqrt{18}\times \frac{4}{27}\sqrt{38}$
$=\frac{18}{2}\times \frac{4}{27}\sqrt{38}=\frac{36}{27}\sqrt{38}$
$=\frac{4}{3}\sqrt{38}$