Question
The plane through the intersection of the planes $$x+y+z=1$$ and $$2x+3y-z+4=0$$ and parallel to $$y$$-axis also passes through the point :
A.
$$\left( { - 3,\,0,\, - 1} \right)$$
B.
$$\left( { - 3,\,1,\,1} \right)$$
C.
$$\left( {3,\,3,\, - 1} \right)$$
D.
$$\left( {3,\,2,\,1} \right)$$
Answer :
$$\left( {3,\,2,\,1} \right)$$
Solution :
Since, equation of plane through intersection of planes
$$x+y+z=1$$ and $$2x+3y-z+4=0$$ is
$$\eqalign{
& \left( {2x + 3y - z + 4} \right) + \lambda \left( {x + y + z - 1} \right) = 0 \cr
& \left( {2 + \lambda } \right)x + \left( {3 + \lambda } \right)y + \left( { - 1 + \lambda } \right)z + \left( {4 - \lambda } \right) = 0.....(1) \cr} $$
But, the above plane is parallel to y-axis then
$$\eqalign{
& \left( {2 + \lambda } \right) \times 0 + \left( {3 + \lambda } \right) \times 1 + \left( { - 1 + \lambda } \right) \times 0 = 0 \cr
& \Rightarrow \lambda = - 3 \cr} $$
Hence, the equation of required plane is
$$\eqalign{
& - x - 4z + 7 = 0 \cr
& \Rightarrow x + 4z - 7 = 0 \cr} $$
Therefore, (3, 2, 1) the passes through the point.