Question
The locus of a point, such that the sum of the squares of its distances from the planes $$x + y + z = 0,\,x - z = 0$$ and $$x - 2y + z = 0$$ is $$9$$, is :
A.
$${x^2} + {y^2} + {z^2} = 3$$
B.
$${x^2} + {y^2} + {z^2} = 6$$
C.
$${x^2} + {y^2} + {z^2} = 9$$
D.
$${x^2} + {y^2} + {z^2} = 12$$
Answer :
$${x^2} + {y^2} + {z^2} = 9$$
Solution :
Let the variable point be $$\left( {\alpha ,\,\beta ,\,\gamma } \right)$$ then according to question
$$\eqalign{
& {\left( {\frac{{\left| {\alpha + \beta + \gamma } \right|}}{{\sqrt 3 }}} \right)^2} + {\left( {\frac{{\left| {\alpha - \gamma } \right|}}{{\sqrt 2 }}} \right)^2} + {\left( {\frac{{\left| {\alpha - 2\beta + \gamma } \right|}}{{\sqrt 6 }}} \right)^2} = 9 \cr
& \Rightarrow {\alpha ^2} + {\beta ^2} + {\gamma ^2} = 9 \cr} $$
So, the locus of the point is $${x^2} + {y^2} + {z^2} = 9$$