Question

If two distinct point $Q , R$ lie on the line of intersection of the planes $- x +2 y - z =0$ and $3 x-5 y+2 z=0$ and $P Q=P R=\sqrt{18}$ 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)

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