If $(\mathrm{x}, \mathrm{y}, \mathrm{z})$ be an arbitrary point lying on a plane $\mathrm{P}$ which passes through the point $(42,0,0)(0,42,0)$ and $(0,0,42)$, then the value of expression
$3+\frac{x-11}{(y-19)^{2}(z-12)^{2}}+\frac{y-19}{(x-11)^{2}(z-12)^{2}}$
$+\frac{z-12}{(x-11)^{2}(y-19)^{2}}-\frac{x+y+z}{14(x-11)(y-19)(z-12)}$
Correct Option: , 2
Plane passing through $(42,0,0),(0,42,0)$,
$(0,0,42)$
From intercept from, equation of plane is
$x+y+z=42$
$\Rightarrow(\mathrm{x}-11)+(\mathrm{y}-19)+(\mathrm{z}-12)=0$
let $\quad a=x-11, b=y-19, c=z-12$
$a+b+c=0$
Now, given expression is
$3+\frac{a}{b^{2} c^{2}}+\frac{b}{a^{2} c^{2}}+\frac{c}{a^{2} b^{2}}-\frac{42}{14 a b c}$
$3+\frac{a^{3}+b^{3}+c^{3}-3 a b c}{a^{2} b^{2} c^{2}}$
If $a+b+c=0$
$\Rightarrow a^{3}+b^{3}+c^{3}=3 a b c$
$\Rightarrow 3$
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