Question
A variable plane which remains at a constant distance $$3p$$ from the origin cut the coordinate axes at $$A,\,B$$ and $$C$$. The locus of the centroid of triangle $$ABC$$ is :
A.
$${x^{ - 1}} + {y^{ - 1}} + {z^{ - 1}} = {p^{ - 1}}$$
B.
$${x^{ - 2}} + {y^{ - 2}} + {z^{ - 2}} = {p^{ - 2}}$$
C.
$$x + y + z = p$$
D.
$${x^2} + {y^2} + {z^2} = {p^2}$$
Answer :
$${x^{ - 2}} + {y^{ - 2}} + {z^{ - 2}} = {p^{ - 2}}$$
Solution :
Let equation of the variable plane be
$$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1......\left( 1 \right)$$
This meets the coordinate axes at $$A\left( {a,\,0,\,0} \right),\,B\left( {0,\,b,\,0} \right)$$ and $$C\left( {0,\,0,\,c} \right).$$
Let $$P\left( {\alpha ,\,\beta ,\,\gamma } \right)$$ be the centroid of the $$\Delta ABC.$$ Then
$$\eqalign{
& \alpha = \frac{{a + 0 + 0}}{3},\,\beta = \frac{{0 + b + 0}}{3},\,\gamma = \frac{{0 + 0 + c}}{3} \cr
& \therefore \,a = 3\alpha ,\,b = 3\beta ,\,c = 3\gamma ......\left( 2 \right) \cr} $$
Plane $$\left( 1 \right)$$ is at constant distance $$3p$$ from the origin, so
$$\eqalign{
& 3p = \frac{{\left| {\frac{0}{a} + \frac{0}{b} + \frac{0}{c} - 1} \right|}}{{\sqrt {{{\left( {\frac{1}{a}} \right)}^2} + {{\left( {\frac{1}{b}} \right)}^2} + {{\left( {\frac{1}{c}} \right)}^2}} }} \cr
& \Rightarrow \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} = \frac{1}{{9{p^2}}}......\left( 3 \right) \cr} $$
From $$\left( 2 \right)$$ and $$\left( 3 \right),$$ we get
$$\eqalign{
& \frac{1}{{9{\alpha ^2}}} + \frac{1}{{9{\beta ^2}}} + \frac{1}{{9{\gamma ^2}}} = \frac{1}{{9{p^2}}} \cr
& \Rightarrow {\alpha ^{ - 2}} + {\beta ^{ - 2}} + {\gamma ^{ - 2}} = {p^{ - 2}} \cr} $$
Generalizing $$\alpha ,\,\beta ,\,\gamma ,$$ locus of centroid $$P\left( {\alpha ,\,\beta ,\,\gamma } \right)$$ is $${x^{ - 2}} + {y^{ - 2}} + {z^{ - 2}} = {p^{ - 2}}$$