Question
The vertices B and C of a triangle ABC lie on the line $\frac{x}{1} = \frac{1 - y}{- 2} = \frac{z - 2}{3}$. The coordinates of A and B are $(1,6,3)$ and $(4,9,\alpha)$ respectively and C is at a distance of 10 units from B . The area (in sq. units) of $\bigtriangleup ABC$ is:
Options
Solution
$\frac{4}{1} = \frac{9 - 1}{2} = \frac{\alpha - 2}{3} \Rightarrow \alpha = 14$
$$\begin{matrix} & \overrightarrow{AD} \cdot (\widehat{i} + 2\widehat{j} + 3\widehat{k}) = 0 \\ & \ (\lambda - 1)\widehat{i} + (2\lambda - 5)\widehat{j} + (3\lambda - 1)\widehat{k} = \overrightarrow{AD} \\ & \ \Rightarrow \lambda - 1 + 4\lambda - 10 + 9\lambda - 3 = 0 \\ & \ \Rightarrow 14\lambda = 14 \Rightarrow \lambda = 1 \\ & D = (1,3,5) \\ & AD = \sqrt{3^{2} + 2^{2}} = \sqrt{13} \\ & Ar( \bigtriangleup ABC) = \frac{1}{2} \times \sqrt{13} \times 10 = 5\sqrt{13} \end{matrix}$$
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