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(B) The equation of the plane is $x + 2y + z = 14$.The length of the perpendicular $PQ$ from point $P(1, 2, 3)$ to the plane is given by:$PQ = \left|

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(B) The equation of the plane is $x + 2y + z = 14$.The length of the perpendicular $PQ$ from point $P(1, 2, 3)$ to the plane is given by:$PQ = \left| \frac{1(1) + 2(2) + 1(3) - 14}{\sqrt{1^2 + 2^2 + 1^2}} ight| = \left| \frac{1 + 4 + 3 - 14}{\sqrt{6}} ight| = \left| \frac{-6}{\sqrt{6}} ight| = \sqrt{6}$.In the right-angled triangle $	riangle PQR$,where $\angle PQR = 90^{\circ}$ and $\angle PRQ = 60^{\circ}$,we have:$QR = PQ \cot(60^{\circ}) = \sqrt{6} 	imes \frac{1}{\sqrt{3}} = \sqrt{2}$.The area of $	riangle PQR$ is given by:Area $= \frac{1}{2} 	imes PQ 	imes QR = \frac{1}{2} 	imes \sqrt{6} 	imes \sqrt{2} = \frac{1}{2} 	imes \sqrt{12} = \frac{1}{2} 	imes 2\sqrt{3} = \sqrt{3}$.

Let $Q$ be the foot of the perpendicular drawn from the point $P(1, 2, 3)$ to the plane $x + 2y + z = 14$. If $R$ is a point on the plane such that $\angle PRQ = 60^{\circ}$,then the area of $\triangle PQR$ is equal to:

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