Question

A point moves such that the sum of square of it's distances from the planes $x-z=0,x-2y+z=0$ and $x+y+z=0$ is $36.$ Then the locus of the point is

• A. $x^2+y^2+z^2=6$
• B. $x^2+y^2+z^2=12$
• C. $x^2+y^2+z^2=18$
• D. $x^2+y^2+z^2=36$

Answer 1

The correct option is D $x^2+y^2+z^2=36$

Given planes are,
$x-z=0,x-2y+z=0$ and $x+y+z=0$
Let moving point is $P(a,b,c)$
$\therefore \frac{|a-c{|}^2}{2}+\frac{|a-2b+c{|}^2}{6}+\frac{|a+b+c{|}^2}{3}=36$
On solving we get,
$a^2+b^2+c^2=36$
Hence, locus of point is : $x^2+y^2+z^2=36$