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

$\begin{array}{l}-x+2y-z=0....\left(1\right) 3x-5y+2z=0.......\left(2\right)\\ \begin{array}{l}\mathrm{drs} \mathrm{of} \mathrm{line} \mathrm{of} \mathrm{intersection} \mathrm{of} \left(1\right) \mathrm{and} \left(2\right) \mathrm{is} \overrightarrow{\mathrm{n}}=\left|\begin{array}{ccc}\widehat{\mathrm{i}}& \widehat{\mathrm{j}}& \widehat{\mathrm{k}}\\ -1& 2& -1\\ 3& -5& 2\end{array}\right|\\ =\widehat{\mathrm{i}}(-1)-\widehat{\mathrm{j}}\left(1\right)+\widehat{\mathrm{k}}(-1)\Rightarrow \overrightarrow{\mathrm{n}}=-\widehat{\mathrm{i}}-\widehat{\mathrm{j}}-\widehat{\mathrm{k}}\\ \mathrm{Equation} \mathrm{of} \mathrm{line} \mathrm{of} \mathrm{intersection} \mathrm{of} \left(1\right) \mathrm{and} \left(2\right) \mathrm{is}\\ \frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{1}=\frac{\mathrm{Z}}{1}\\ \mathrm{Let} \mathrm{T}(\mathrm{\alpha },\mathrm{\alpha },\mathrm{\alpha }) \mathrm{be} \mathrm{the} \mathrm{foot} \mathrm{of} \mathrm{perpendicular} \mathrm{of} \mathrm{P} \mathrm{on} \mathrm{the} \mathrm{line} \mathrm{QR}\\ \mathrm{drs}:\text{ of PT }\to \alpha -1,\alpha +2,\alpha -3 and \\ \text{ drs: of }\mathrm{QR}\to 1,1,1,\\ \begin{array}{l}PT\perp QR\Rightarrow (\alpha -1)\times 1+(\alpha +2)\times 1+(\alpha -3)\times 1=0\\ 3\alpha =2\Rightarrow \alpha =\frac{2}{3}\therefore T\left(\frac{2}{3},\frac{2}{3},\frac{2}{3}\right),P\left(1,-2,3\right)\\ \\ {\mathrm{PT}}^2=\frac{1}{9}+\frac{64}{9}+\frac{49}{9}\Rightarrow {\mathrm{PT}}^2=\frac{114}{9}\therefore \mathrm{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}\\ \cos 2\theta =\frac{2\times 19}{27}-1=\frac{11}{27}\Rightarrow \\ \sin 2\theta =\sqrt{1-{\left(\frac{11}{27}\right)}^2}=\frac{\sqrt{38}\sqrt{16}}{27}==\frac{4}{27}\sqrt{38}\\ \begin{array}{c} \\ \text{ Area of ∆PQR is }=\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}\end{array}\end{array}\end{array}\end{array}$