Question
If lines $$x = y = z$$ and $$x = \frac{y}{2} = \frac{z}{3}$$ and third line passing through $$\left( {1,\,1,\,1} \right)$$ form a triangle of area $$\sqrt 6 $$ units, then the point of intersection of third line with the second line will be :
A.
$$\left( {1,\,2,\,3} \right)$$
B.
$$\left( {2,\,4,\,6} \right)$$
C.
$$\left( {\frac{4}{3},\,\frac{8}{3},\,\frac{{12}}{3}} \right)$$
D.
None of these
Answer :
$$\left( {2,\,4,\,6} \right)$$
Solution :
$${\text{Let any point on the second line be}}\left( {\lambda ,\,2\lambda ,\,3\lambda } \right)$$
$$\eqalign{ & \cos \,\theta = \frac{6}{{\sqrt {42} }},\,\sin \,\theta = \frac{{\sqrt 6 }}{{\sqrt {42} }} \cr & {\Delta _{OAB}} = \frac{1}{2}\left( {OA} \right)OB\,\sin \,\theta \cr & = \frac{1}{2}\sqrt 3 \lambda \sqrt {14} \times \frac{{\sqrt 6 }}{{\sqrt {42} }} \cr & = \sqrt 6 {\text{ or }}\lambda = 2 \cr & {\text{So, }}B{\text{ is }}\left( {2,\,4,\,6} \right) \cr} $$
$${\text{Let any point on the second line be}}\left( {\lambda ,\,2\lambda ,\,3\lambda } \right)$$
$$\eqalign{ & \cos \,\theta = \frac{6}{{\sqrt {42} }},\,\sin \,\theta = \frac{{\sqrt 6 }}{{\sqrt {42} }} \cr & {\Delta _{OAB}} = \frac{1}{2}\left( {OA} \right)OB\,\sin \,\theta \cr & = \frac{1}{2}\sqrt 3 \lambda \sqrt {14} \times \frac{{\sqrt 6 }}{{\sqrt {42} }} \cr & = \sqrt 6 {\text{ or }}\lambda = 2 \cr & {\text{So, }}B{\text{ is }}\left( {2,\,4,\,6} \right) \cr} $$