Question
The equation of a plane passing through the line of intersection of the planes $$x+2y+3z=2$$ and $$x-y+z=3$$ and at a distance $$\frac{2}{{\sqrt 3 }}$$ from the point $$\left( {3,\,1,\, - 1} \right)$$ is :
A.
$$5x-11y+z=17$$
B.
$$\sqrt 2 x + y = 3\sqrt 2 - 1$$
C.
$$x + y + z = \sqrt 3 $$
D.
$$x - \sqrt 2 y = 1 - \sqrt 2 $$
Answer :
$$5x-11y+z=17$$
Solution :
The plane passing through the intersection line of given planes is
$$\eqalign{
& \left( {x + 2y + 3z - 2} \right) + \lambda \left( {x - y + z - 3} \right) = 0 \cr
& {\text{or}}\,\,\,\left( {1 + \lambda } \right)x + \left( {2 - \lambda } \right)y + \left( {3 + \lambda } \right)z + \left( { - 2 - 3\lambda } \right) = 0 \cr} $$
Its distance from the point $$\left( {3,\,1,\, - 1} \right)$$ is $$\frac{2}{{\sqrt 3 }}$$
$$\eqalign{
& \therefore \left| {\frac{{3\left( {1 + \lambda } \right) + 1\left( {2 - \lambda } \right) - 1\left( {3 + \lambda } \right) + \left( { - 2 - 3\lambda } \right)}}{{\sqrt {{{\left( {1 + \lambda } \right)}^2} + {{\left( {2 - \lambda } \right)}^2} + {{\left( {3 + \lambda } \right)}^2}} }}} \right| = \frac{2}{{\sqrt 3 }} \cr
& \Rightarrow \left| {\frac{{ - 2\lambda }}{{\sqrt {3{\lambda ^2} + 4\lambda + 14} }}} \right| = \frac{2}{{\sqrt 3 }} \cr
& \Rightarrow 3{\lambda ^2} + 4\lambda + 14 = 3{\lambda ^2} \cr
& \Rightarrow \lambda = - \frac{7}{2} \cr} $$
$$\therefore $$ Required equation of plane is
$$\eqalign{
& \left( {x + 2y + 3z - 2} \right) - \frac{7}{2}\left( {x - y + z - 3} \right) = 0 \cr
& {\text{or }}5x - 11y + z = 17 \cr} $$