JEE Main 2021 (Online) 16th March Evening ShiftQuestion 1
If $(x,y,z)$ be an arbitrary point lying on a plane $P$ which passes through the points $(42,0,0)$, $(0,42,0)$ and $(0,0,42)$, then the value of the 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 equal to :
3D Geometry
Solution
Put $a=x-11,b=y-19,c=z-12$; then $a+b+c=42-42=0$ and $x+y+z=42$. Expression $=3+\frac{a^3+b^3+c^3}{a^2b^2c^2}-\frac{3}{abc}=3+\frac{3abc}{a^2b^2c^2}-\frac3{abc}=3$.