# Let P be an arbitrary point having sum

Question:

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

Solution:

Let point $\mathrm{P}$ is $(\alpha, \beta, \gamma)$

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

Locus is

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

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

Since its given that $x^{2}+y^{2}+z^{2}=9$ After solving $\ell=\mathrm{n}$

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