Let P be an arbitrary point such that the sum | vedclass

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(A) Let the point $P$ be $(x, y, z)$. The distances from the planes $x + y + z = 0$,$lx - nz = 0$,and $x - 2y + z = 0$ are given by $d_1 = \frac{|x +

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(A) Let the point $P$ be $(x, y, z)$. The distances from the planes $x + y + z = 0$,$lx - nz = 0$,and $x - 2y + z = 0$ are given by $d_1 = \frac{|x + y + z|}{\sqrt{3}}$,$d_2 = \frac{|lx - nz|}{\sqrt{l^2 + n^2}}$,and $d_3 = \frac{|x - 2y + z|}{\sqrt{6}}$.Given $d_1^2 + d_2^2 + d_3^2 = 9$,we have:$\frac{(x + y + z)^2}{3} + \frac{(lx - nz)^2}{l^2 + n^2} + \frac{(x - 2y + z)^2}{6} = 9$.Expanding the terms:$\frac{x^2 + y^2 + z^2 + 2xy + 2yz + 2zx}{3} + \frac{l^2x^2 - 2lnxz + n^2z^2}{l^2 + n^2} + \frac{x^2 + 4y^2 + z^2 - 4xy - 4yz + 2zx}{6} = 9$.Grouping the coefficients of $x^2, y^2, z^2, xy, yz, zx$:$x^2(\frac{1}{3} + \frac{l^2}{l^2 + n^2} + \frac{1}{6}) + y^2(\frac{1}{3} + \frac{4}{6}) + z^2(\frac{1}{3} + \frac{n^2}{l^2 + n^2} + \frac{1}{6}) + xy(\frac{2}{3} - \frac{4}{6}) + yz(\frac{2}{3} - \frac{4}{6}) + zx(\frac{2}{3} - \frac{2ln}{l^2 + n^2} + \frac{2}{6}) = 9$.Simplifying coefficients:$x^2(\frac{1}{2} + \frac{l^2}{l^2 + n^2}) + y^2(1) + z^2(\frac{1}{2} + \frac{n^2}{l^2 + n^2}) + zx(1 - \frac{2ln}{l^2 + n^2}) = 9$.Comparing this with $x^2 + y^2 + z^2 = 9$,the coefficients of $xy, yz, zx$ must be $0$ and the coefficients of $x^2, y^2, z^2$ must be $1$.For the coefficient of $zx$ to be $0$,$1 - \frac{2ln}{l^2 + n^2} = 0 \implies l^2 + n^2 = 2\ln \implies (l - n)^2 = 0 \implies l = n$.Thus,$l - n = 0$.

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

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