Question
A variable plane at a distance of $$1$$ unit from the origin cuts the coordinate axes at $$A,\,B$$ and $$C.$$ If the centroid $$D\left( {x,\,y,\,z} \right)$$ satisfies the relation $${x^{ - 2}} + {y^{ - 2}} + {z^{ - 2}} = k$$ then the value of $$k$$ is :
A.
$$3$$
B.
$$1$$
C.
$$\frac{1}{3}$$
D.
$$9$$
Answer :
$$9$$
Solution :
The plane is $$lx + my + nz = 1$$ where $${l^2} + {m^2} + {n^2} = 1.$$ It cuts axes at $$\left( {\frac{1}{l},\,0,\,0} \right),\,\left( {0,\,\frac{1}{m},\,0} \right),\,\left( {0,\,0,\,\frac{1}{n}} \right).$$
$$\therefore $$ the centroid $$ = \left( {\frac{1}{{3l}},\,\frac{1}{{3m}},\,\frac{1}{{3n}}} \right).$$ It satisfies $$\frac{1}{{{x^2}}} + \frac{1}{{{y^2}}} + \frac{1}{{{z^2}}} = k\,\,\, \Rightarrow 9\left( {{l^2} + {m^2} + {n^2}} \right) = k\,\,\, \Rightarrow 9 = k.$$