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

Given $Q$ be the foot of perpendicular drawn from the point $P\left(1,2,3\right)$ to the plane $x+2y+z=14$. If $R$ is a point on the plane such that $\angle PRQ=60°$, then the area of $∆PQR$ is equal to

Now plotting the diagram of given value we get,

 

Now given plane $x+2y+z=14$

Length of perpendicular from $P \text{to} Q$ will be

$PQ=\left|\frac{1+4+3-14}{\sqrt{6}}\right|=\sqrt{6}$

Now in $△PQR$

$QR=\left(PQ\right)\cot 60°=\sqrt{2}$

So,  Area of $∆PQR=\frac{1}{2}\left(PQ\right)\left(QR\right)=\frac{1}{2}\times \sqrt{6}\times \sqrt{2}=\sqrt{3}$