Let $$P$$ be an arbitrary point having sum of the squares of the distances 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 ________.

Let $$P = (x, y, z)$$. The three planes are $$\pi_1: x + y + z = 0$$, $$\pi_2: lx - nz = 0$$, and $$\pi_3: x - 2y + z = 0$$.

The squared distances from $$P$$ to these planes are $$d_1^2 = \frac{(x+y+z)^2}{3}$$, $$d_2^2 = \frac{(lx - nz)^2}{l^2 + n^2}$$, and $$d_3^2 = \frac{(x - 2y + z)^2}{6}$$.

The condition $$d_1^2 + d_2^2 + d_3^2 = 9$$ must give the locus $$x^2 + y^2 + z^2 = 9$$, so the left side must be identically equal to $$x^2 + y^2 + z^2$$.

Expanding the sum of the first and third terms: $$\frac{(x+y+z)^2}{3} + \frac{(x-2y+z)^2}{6}$$. Computing the numerator over the common denominator 6: $$\frac{2(x+y+z)^2 + (x-2y+z)^2}{6}$$.

Expanding: $$2(x+y+z)^2 = 2x^2 + 2y^2 + 2z^2 + 4xy + 4yz + 4xz$$ and $$(x-2y+z)^2 = x^2 + 4y^2 + z^2 - 4xy - 4yz + 2xz$$. Their sum is $$3x^2 + 6y^2 + 3z^2 + 6xz = 3(x+z)^2 + 6y^2$$.

So the sum of the first and third terms is $$\frac{(x+z)^2 + 2y^2}{2}$$. For the identity to hold, the second term must supply the remainder: $$\frac{(lx-nz)^2}{l^2+n^2} = x^2+y^2+z^2 - \frac{(x+z)^2+2y^2}{2} = \frac{2x^2+2y^2+2z^2 - x^2 - 2xz - z^2 - 2y^2}{2} = \frac{x^2 - 2xz + z^2}{2} = \frac{(x-z)^2}{2}$$.

Comparing $$\frac{(lx-nz)^2}{l^2+n^2} = \frac{(x-z)^2}{2}$$ as an identity in $$x$$ and $$z$$, equating the coefficient of $$x^2$$: $$\frac{l^2}{l^2+n^2} = \frac{1}{2}$$, so $$l^2 = n^2$$. Equating the coefficient of $$xz$$: $$\frac{-2ln}{l^2+n^2} = \frac{-2}{2} = -1$$, so $$2ln = l^2+n^2$$. Since $$l^2 = n^2$$, this gives $$2ln = 2l^2$$, hence $$n = l$$.

Therefore $$l - n = 0$$.