Let P be the foot of the perpendicular from the point Q(10, –3, –1) on the line $\frac{x - 3}{7} = \frac{y - 2}{- 1} = \frac{z + 1}{- 2} .$ Then the area of the right angled triangle PQR, where R is the point (3, –2, 1), is

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8 $\sqrt{15}$

3 $\sqrt{30}$

$\sqrt{30}$

9 $\sqrt{15}$

answer is D.

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Detailed Solution

$\begin{aligned} \begin{aligned} \frac{x - 3}{7} = \frac{y - 2}{- 1} = \frac{z + 1}{- 2} = λ \\ \Rightarrow 7 λ + 3, - λ + 2, - 2 λ - 1 \\ dr's of QP \Rightarrow \\ 7 λ - 7, - λ + 5, - 2 λ \end{aligned} \\ Now \\ \begin{aligned} (7 λ - 7) \cdot 7 - (- λ + 5) + (2 λ) \cdot 2 = 0 \\ 54 λ - 54 = 0 \Rightarrow λ = 1 \\ ∴ P = (10, 1, - 3) \\ \vec{PQ} = - 4 \hat{j} + 2 \hat{k} \\ \vec{PR} = - 7 \hat{i} - 3 \hat{j} + 4 \hat{k} \\ Area = |\frac{1}{2}| \begin{array}{ccc} i & j & k \\ 0 & - 4 & 2 \\ - 7 & - 3 & 4 \end{array} | | = 3 \sqrt{30} \end{aligned} \end{aligned}$