Question

Let a point $P$ moves such that the sum of squares of it's distance from the planes $P_1:x-y-z=0$ and $P_2:y-z=0$ is always equal to the square of it's distance from the plane $P_3:x-2y+z=0.$ Then the locus of point $P$ is:

• A. $2x^2+y^2-4z^2+4xy+2yz-6xz=0$
• B. $x^2+y^2+4z^2+2yz-6xz=0$
• C. $2x^2+3y^2+4z^2+4xz-6xy=0$
• D. $x^2-4y^2+z^2+2yz-6xz=0$

Answer 1

The correct option is B $x^2+y^2+4z^2+2yz-6xz=0$

Let the point $P\equiv (p,q,r)$
$P_1:x-y-z=0$ and $P_2:y-z=0,$ $P_3:x-2y+z=0.$
We know, distance of a point $(x_1,y_1,z_1)$ from plane $ax+by+cz=d$ is $\frac{|a{x}_1+b{y}_1+c{z}_1-d|}{\sqrt{a^2+b^2+c^2}}$
According to statement :
$\frac{(p-q-r{)}^2}{3}+\frac{(q-r{)}^2}{2}=\frac{(p-2q+r{)}^2}{6}$
on solving, we get
$\Rightarrow p^2+q^2+4r^2+2qr-6pr=0$
So, locus of point is :
$x^2+y^2+4z^2+2yz-6xz=0$