Let the image of the point P(1,2,3) in the plane 2x–y+z=9 be Q. If the coordinates of the point R are (6,10,7), then the square of the area of the…

Given,

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

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

Now plotting the diagram we get,

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

Now length of $PR=\sqrt{{5}^{2}+{8}^{2}+{4}^{2}}=\sqrt{105}$PR=52+82+42​=105​

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

$$\mathrm{cos}\theta =\frac{\vec{PR}\cdot \vec{PM}}{|\vec{PR}|\cdot |\vec{PM}|}$$cosθ=∣PR∣⋅∣PM∣PR⋅PM​

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

$$\Rightarrow \mathrm{cos}\theta =\frac{6}{\sqrt{630}}$$⇒cosθ=630​6​

Now Area$(\Delta PQR)=2area(\Delta PMR)$(ΔPQR)=2area(ΔPMR)

$$\Rightarrow \Delta PQR=2\cdot \frac{1}{2}\cdot PR\mathrm{sin}\theta \cdot PR\mathrm{cos}\theta$$⇒ΔPQR=2⋅21​⋅PRsinθ⋅PRcosθ

$$\Rightarrow \Delta PQR=2\cdot \frac{1}{2}(PR{)}^{2}\mathrm{sin}\theta \mathrm{cos}\theta$$⇒ΔPQR=2⋅21​(PR)2sinθcosθ

$$\Rightarrow \Delta PQR=105\cdot \frac{6}{\sqrt{630}}\cdot \frac{\sqrt{594}}{\sqrt{630}}$$⇒ΔPQR=105⋅630​6​⋅630​594​​

$$\Rightarrow \Delta PQR=\sqrt{594}$$⇒ΔPQR=594​

Hence, area of the square will be $594$594