Question
A variable plane passes through a fixed point $$\left( {1,\,2,\,3} \right).$$ The locus of the foot of the perpendicular from the origin to this plane is given by :
A.
$${x^2} + {y^2} + {z^2} - 14 = 0$$
B.
$${x^2} + {y^2} + {z^2} + x + 2y + 3z = 0$$
C.
$${x^2} + {y^2} + {z^2} - x - 2y - 3z = 0$$
D.
None of these
Answer :
$${x^2} + {y^2} + {z^2} - x - 2y - 3z = 0$$
Solution :
Let $$P\left( {\alpha ,\,\beta ,\,\gamma } \right)$$ be the foot of the perpendicular from the origin $$O\left( {0,\,0,\,0} \right)$$ to the plane So, the plane passes through $$P\left( {\alpha ,\,\beta ,\,\gamma } \right)$$ and is perpendicular to $$OP.$$ Clearly direction ratios of $$OP$$ i.e., normal to the plane are $${\alpha ,\,\beta ,\,\gamma }.$$ Therefore, equation of the plane is
$$\alpha \left( {x - \alpha } \right) + \beta \left( {y - \beta } \right) + \gamma \left( {z - \gamma } \right) = 0$$
This plane passes through the fixed point $$\left( {1,\,2,\,3} \right),$$ so
$$\eqalign{
& \alpha \left( {1 - \alpha } \right) + \beta \left( {2 - \beta } \right) + \gamma \left( {3 - \gamma } \right) = 0 \cr
& {\text{or, }}{\alpha ^2} + {\beta ^2} + {\gamma ^2} - \alpha - 2\beta - 3\gamma = 0 \cr} $$
Generalizing $${\alpha ,\,\beta }$$ and $$\gamma $$, locus of $$P\left( {\alpha ,\,\beta ,\,\gamma } \right)$$ is
$${x^2} + {y^2} + {z^2} - x - 2y - 3z = 0$$