Let the image of the point P ( 1 ,   2 ,   3 ) in the plane 2 x – y + z = 9 be Q . If the…

Given,

The image of the point $P(1, 2, 3)$ in the plane $2x–y+z=9$ be $Q$,

The coordinates of the point $R$ are $(6, 10, 7),$ 

Now plotting the diagram we get,

$R$ lies on plane as $\left(6,10,7\right)$ satisfy the plane equation $2x–y+z=9$

Now length of $PR=\sqrt{5^2+8^2+4^2}=\sqrt{105}$

Now finding the angle between $PR \& PM$(which is normal vector to plane) we get,

$\cos \theta =\frac{\overrightarrow{PR}·\overrightarrow{PM}}{\left|\overrightarrow{PR}\right|·\left|\overrightarrow{PM}\right|}$

$\Rightarrow \cos \theta =\frac{(5\hat{\mathrm{i}}+8\hat{\mathrm{j}}+4\hat{\mathrm{k}})(2\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})}{\sqrt{105}\sqrt{6}}$

$\Rightarrow \cos \theta =\frac{6}{\sqrt{630}}$

Now Area$(∆PQR)=2area(∆PMR)$

$\Rightarrow ∆PQR=2·\frac{1}{2}·PR\sin \theta ·PR\cos \theta$

$\Rightarrow ∆PQR=2·\frac{1}{2}(PR{)}^2\sin \theta \cos \theta$

$\Rightarrow ∆PQR=105·\frac{6}{\sqrt{630}}·\frac{\sqrt{594}}{\sqrt{630}}$

$\Rightarrow ∆PQR=\sqrt{594}$

Hence, area of the square will be $594$