If $$(x, y, z)$$ be an arbitrary point lying on a plane $$P$$ which passes through the point $$(42, 0, 0)$$, $$(0, 42, 0)$$ and $$(0, 0, 42)$$, then the value of expression $$3 + \frac{x-11}{(y-19)^2(z-12)^2} + \frac{y-19}{(x-11)^2(z-12)^2} + \frac{z-12}{(x-11)^2(y-19)^2} - \frac{x+y+z}{14(x-11)(y-19)(z-12)}$$ is

The plane passes through $$(42, 0, 0)$$, $$(0, 42, 0)$$, and $$(0, 0, 42)$$, so its equation is $$x + y + z = 42$$.

Let $$a = x - 11$$, $$b = y - 19$$, $$c = z - 12$$. Since $$x + y + z = 42$$, we get $$a + b + c = (x + y + z) - 42 = 0$$.

The expression becomes: $$3 + \frac{a}{b^2c^2} + \frac{b}{a^2c^2} + \frac{c}{a^2b^2} - \frac{x+y+z}{14abc}$$.

Since $$x + y + z = 42$$, the last term is $$\frac{42}{14abc} = \frac{3}{abc}$$.

Combining the three fraction terms over the common denominator $$a^2b^2c^2$$: $$\frac{a^3 + b^3 + c^3}{a^2b^2c^2}$$.

Now we use the identity: when $$a + b + c = 0$$, we have $$a^3 + b^3 + c^3 = 3abc$$.

So the expression becomes: $$3 + \frac{3abc}{a^2b^2c^2} - \frac{3}{abc} = 3 + \frac{3}{abc} - \frac{3}{abc} = 3$$.