Question

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 :

• A. $3$
• B. $0$
• C. $39$
• D. $-45$

Answer 1

The correct option is A $3$

Equation of plane is $x+y+z=42$
or, $(x-11)+(y-19)+(z-12)=0$
Now, $\frac{x-11}{(y-19{)}^2(z-12{)}^2}+\frac{y-19}{(x-11{)}^2(z-12{)}^2}+\frac{z-12}{(z-11{)}^2(y-19{)}^2}$
$=\frac{(x-11{)}^3+(y-19{)}^3+(z-12{)}^3}{(x-11{)}^2(y-19{)}^2(z-12{)}^2}$
$=\frac{3(x-11)(y-19)(z-12)}{(x-11{)}^2(y-19{)}^2(z-12{)}^2}$
$[\text{If}a+b+c=0,\text{then}{a}^3+b^3+c^3=3abc]$
$=\frac{3}{(x-11)(y-19)(z-12)}$
$\therefore$ The given expression is equal to
$3+\frac{3}{(x-11)(y-19)(z-12)}-\frac{3}{(x-11)(y-19)(z-12)}=3$