Question

Let Q be the foot of 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 ∠PRQ = 60°, then the area of ΔPQR is equal to :

• A. $\sqrt{\frac{3}{2}}$
• B. $\sqrt3$
• C. $2\sqrt3$
• D. 3

Answer 1

Answer: (B)

$\sqrt3$

Answer 2

PQ=|$\frac{1+4+3−14}{\sqrt6}$|=√6
QR=$\frac{PQ}{tan60^{\circ}}$=$\frac{\sqrt6}{\sqrt3}$=$\sqrt2$
Area(ΔPQR)=$\frac{1}{2}$⋅PQ⋅QR=$\sqrt3$
So, the correct option is (B): $\sqrt3$