question_answer

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

A) \[(1,2,3)\]

B) \[(2,4,6)\]

C) \[\left( \frac{4}{3},\frac{8}{3},\frac{12}{3} \right)\]

D) None of these

Correct Answer: B

Solution :

[b] Let any point on the second line be \[(\lambda ,2\lambda ,3\lambda )\] \[\cos \theta =\frac{6}{\sqrt{42}},\sin \theta =\frac{\sqrt{6}}{\sqrt{42}}\] \[{{\Delta }_{OAB}}=\frac{1}{2}(OA)\,\,OB\,\,sin\theta \] \[=\frac{1}{2}\sqrt{3}\lambda \sqrt{14}\times \frac{\sqrt{6}}{\sqrt{42}}=\sqrt{6}or\lambda =2\] So, B is (2, 4, 6)