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  • Let be the foot of perpendicular drawn from the point <img alt="\mathrm{P(1,2,3) }" src="/latex-image/?%5Cmathrm%7BP%281%2C2%2C3%29%20%

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

Option: 1

\frac{\sqrt{3}}{2}


Option: 2

\sqrt{3}


Option: 3

2 \sqrt{3}


Option: 4

3


Answers (1)

best_answer

\begin{gathered} \mathrm{P Q=\frac{11+4+3-14 \mid}{\sqrt{1+4+1}}=\frac{6}{\sqrt{6}}=\sqrt{6}} \\ \text { Area }\mathrm{=\frac{1}{2} \cdot P Q \cdot Q R=\frac{1}{2} \cdot \sqrt{6} \times \frac{P Q}{\tan 60^{\circ}} }\\ \mathrm{=\frac{1}{2} \times \frac{6}{\sqrt{3}}=\sqrt{3}} \end{gathered}

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