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

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 $PR \text{ and } PM$(which is normal vector to plane) we get,

$cos\theta = \dfrac{\overset{\to}{PR} \cdot \overset{\to}{PM}}{\left|\overset{\to}{PR}\right| \cdot \left|\overset{\to}{PM}\right|}$

⇒ cos\theta = \dfrac{( 5\overset{^}{i} + 8\overset{^}{j} + 4\overset{^}{k} ) ( 2\overset{^}{i} - \overset{^}{j} + \overset{^}{k} )}{\sqrt{105}\sqrt{6}}

$⇒ cos\theta = \dfrac{6}{\sqrt{630}}$

Now Area$( ∆ PQR ) = 2 area ( ∆ PMR )$

$⇒ ∆ PQR = 2 \cdot \dfrac{1}{2} \cdot PRsin\theta \cdot PRcos\theta$

$⇒ ∆ PQR = 2 \cdot \dfrac{1}{2} ( PR{)}^{2}sin\theta cos\theta$

$⇒ ∆ PQR = 105 \cdot \dfrac{6}{\sqrt{630}} \cdot \dfrac{\sqrt{594}}{\sqrt{630}}$

$⇒ ∆ PQR = \sqrt{594}$

Hence, area of the square will be $594$