Question

If $(x,y,z)$ be an arbitrary point lying on a plane $P$ which passes through the points $(42,0,0),(0,42,0)\mathrm{and}(0,0,42)$, 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

$3$

Explanation for the correct option.

The equation of a plane is $x+y+z=a$. Here, $a=42$. So, the equation of the plane will be

$x+y+z=42......\left(1\right)$

Now, in the given expression $3+\frac{x-11}{{(y-19)}^2{(z-12)}^2}+\frac{y-19}{{(x-11)}^2{(z-12)}^2}+\frac{z-12}{{(x-11)}^2{(y-19)}^2}–\frac{x+y+z}{14(x-11)(y-19)(z-12)}$

Let, $x-11=a,y-19=b\mathrm{and}z-12=c$

So,

$$\begin{gathered} a+b+c=x-11+y-19+z-12 \\ =x+y+z-42 \\ =42-42 \\ =0 \end{gathered}$$

Now, the expression will be,

$$\begin{gathered} 3+\frac{a}{b^2{c}^2}+\frac{b}{a^2{c}^2}+\frac{c}{a^2{b}^2}–\frac{42}{14abc}\left[x+y+z=42\right] \\ =3+\frac{a^3+b^3+c^3-3abc}{a^2{b}^2{c}^2} \\ =3+\frac{3abc-3abc}{a^2{b}^2{c}^2}\left[\mathrm{If}a+b+c=0,\mathrm{then}{a}^3+b^3+c^3=3abc\right] \\ =3 \end{gathered}$$

Hence, option A is correct.