# If (x, y, z) be an arbitrary point lying on a plane

Question:

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

$3+\frac{\mathrm{x}-11}{(\mathrm{y}-19)^{2}(\mathrm{z}-12)^{2}}+\frac{\mathrm{y}-19}{(\mathrm{x}-11)^{2}(\mathrm{z}-12)^{2}}$

$+\frac{\mathrm{z}-12}{(\mathrm{x}-11)^{2}(\mathrm{y}-19)^{2}}-\frac{\mathrm{x}+\mathrm{y}+\mathrm{z}}{14(\mathrm{x}-11)(\mathrm{y}-19)(\mathrm{z}-12)}$

1. 0

2. 3

3. 39

4. -45

Correct Option: , 2

Solution:

Plane passing through $(42,0,0),(0,42,0)$, $(0,0,42)$

From intercept from, equation of plane is

$x+y+z=42$

$\Rightarrow(x-11)+(y-19)+(z-12)=0$

let $\quad a=x-11, b=y-19, c=z-12$

$a+b+c=0$

Now, given expression is

$3+\frac{\mathrm{a}}{\mathrm{b}^{2} \mathrm{c}^{2}}+\frac{\mathrm{b}}{\mathrm{a}^{2} \mathrm{c}^{2}}+\frac{\mathrm{c}}{\mathrm{a}^{2} \mathrm{~b}^{2}}-\frac{42}{14 \mathrm{abc}}$

$3+\frac{a^{3}+b^{3}+c^{3}-3 a b c}{a^{2} b^{2} c^{2}}$

If $a+b+c=0$

$\Rightarrow \mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3}=3 \mathrm{abc}$

$\Rightarrow 3$