Let $P$ be an arbitrary point having sum of the squares of the distance from the planes $x+y+z=0, lx-nz=0$ and $x-2y+z=0$ equal to $9$ units. If the locus of the point $P$ is $x^2+y^2+z^2=9,$ then the value of $l-n$ is equal to

The distance of a point $\left(x_1, y_1, z_1\right)$ from a plane $ax+by+cz+d=0$ is $\frac{\left|a{x}_1+b{y}_1+c{z}_1+d\right|}{\sqrt{a^2+b^2+c^2}}.$

Given the point $P$ is an arbitrary point such that the sum of the squares of its distance from the planes $x+y+z=0, lx-nz=0$ and $x-2y+z=0$ is equal to $9$ units.

Let, the point $P$ is $\left(\alpha , \beta , \gamma \right),$ then we have 

${\left(\frac{\alpha +\beta +\gamma }{\sqrt{3}}\right)}^2+{\left(\frac{l\alpha -\mathrm{n}\gamma }{\sqrt{l^2+{\mathrm{n}}^2}}\right)}^2+{\left(\frac{\alpha -2\beta +\gamma }{\sqrt{6}}\right)}^2=9$

Thus, the locus can be obtained by replacing $\left(\alpha , \beta , \gamma \right)$ by $\left(x, y, z\right)$ and hence the locus is

$\frac{(x+y+z{)}^2}{3}+\frac{(lx-nz{)}^2}{l^2+n^2}+\frac{(x-2y+z{)}^2}{6}=9$

$\Rightarrow \frac{x^2+y^2+z^2+2xy+2yz+2xz}{3}+\frac{l^2{x}^2+n^2{z}^2-2\ln xz}{l^2+n^2}+\frac{x^2+4y^2+z^2-4xy-4yz+2xz}{6}=9$

$\Rightarrow x^2\left(\frac{1}{2}+\frac{l^2}{l^2+n^2}\right)+y^2+z^2\left(\frac{1}{2}+\frac{n^2}{l^2+n^2}\right)+2xz\left(\frac{1}{2}-\frac{nl}{l^2+n^2}\right)-9=0$

Since, it is given that the locus is $x^2+y^2+z^2=9$

$\Rightarrow \frac{1}{2}-\frac{nl}{l^2+n^2}=0$

$\Rightarrow \frac{l^2+n^2-2nl}{2\left(l^2+n^2\right)}=0$

$\Rightarrow {\left(l-n\right)}^2=0$

$\Rightarrow l-n=0.$