If lines $x = y = \mathcal{z}$ and $x = \frac{y}{2} = \frac{\mathcal{z}}{3}$, and third line passing through (1, 1, 1) form a triangle of area $\sqrt{6}$ units, then point of intersection of third line with second line will be

Question

Dr. Abubacker · Accepted Answer

(b)
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Let any point on second line be $(\lambda,\ 2\lambda,\ 3\lambda)$
$\cos	heta = \frac{6}{\sqrt{42}},\sin	heta = \frac{\sqrt{6}}{\sqrt{42}}$
$\Delta_{OAB} = \frac{1}{2}(OA)OB\sin	heta = \frac{1}{2}\sqrt{3}\ \lambda\sqrt{14} 	imes \frac{\sqrt{6}}{\sqrt{42}} = \sqrt{6}$
$\Rightarrow \lambda = 2$
So $B$ is (2, 4, 6)

Q. If lines $x = y = \mathcal{z}$ and $x = \frac{y}{2} = \frac{\mathcal{z}}{3}$, and third line passing through (1, 1, 1) form a triangle of area $\sqrt{6}$ units, then point of intersection of third line with second line will be

Explanation

Reviewed by Dr. Abubacker

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