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 $$\angle PRQ = 60°$$, then the area of $$\Delta PQR$$ is equal to

We have the point $$P(1, 2, 3)$$ and the plane $$x + 2y + z = 14$$. The normal to the plane is $$\vec{n} = \langle 1, 2, 1 \rangle$$, so the foot of the perpendicular $$Q$$ from $$P$$ lies along the line through $$P$$ in the direction of $$\vec{n}$$.

The perpendicular distance from $$P$$ to the plane is $$PQ = \frac{|1 + 2(2) + 3 - 14|}{\sqrt{1^2 + 2^2 + 1^2}} = \frac{|1 + 4 + 3 - 14|}{\sqrt{6}} = \frac{6}{\sqrt{6}} = \sqrt{6}$$.

Since $$Q$$ is the foot of the perpendicular from $$P$$ to the plane, the segment $$PQ$$ is perpendicular to the plane itself. This means $$PQ \perp QR$$ for any point $$R$$ in the plane, giving $$\angle PQR = 90°$$.

In the right triangle $$PQR$$, we have $$\angle PQR = 90°$$ and $$\angle PRQ = 60°$$, so the remaining angle is $$\angle QPR = 30°$$. From the right triangle, $$\tan(\angle PRQ) = \frac{PQ}{QR}$$, so $$\tan 60° = \frac{\sqrt{6}}{QR}$$, giving $$QR = \frac{\sqrt{6}}{\sqrt{3}} = \sqrt{2}$$.

The area of $$\triangle PQR$$ is $$\frac{1}{2} \times PQ \times QR = \frac{1}{2} \times \sqrt{6} \times \sqrt{2} = \frac{1}{2} \times 2\sqrt{3} = \sqrt{3}$$.

Hence, the correct answer is Option B.