Equation of a plane passing through $(-1,1,1)$ and $(1,-1,1)$ and perpendicular to $x+2 y+2 z=5$ is

Equation of plane passing through $(-1,1,1)$ is $a(x+1)+b(y-1)+c(z-1)=0$ ....(i) Also, it is passing through $(1,-1,1)$. $\therefore \quad a(1+1)+b(-1-1)+c(1-1)=0$ $\Rightarrow \quad 2 a-2 b+0 c=0$ ....(ii) Also, required equation of plane (i) is perpendicular to $x+2 y+2 z=5$. $\therefore \quad a \times 1+b \times 2+c \times 2=0$ $\Rightarrow \quad a+2 b+2 c=0$ Eqs. (ii) and (iii) are identical. $\therefore \quad \frac{a}{-4-0}=\frac{-b}{4-0}=\frac{c}{4+2}$ $\Rightarrow \quad \frac{a}{-4}=\frac{b}{-4}=\frac{c}{6}$ $\Rightarrow \quad \frac{a}{-2}=\frac{b}{-2}=\frac{c}{3}=\lambda$ (say) $\Rightarrow \quad a=-2 \lambda, b=-2 \lambda, c=3 \lambda$ On putting the values of $a, b$ and $c$ in Eq. (i), we get $-2 \lambda(x+1)-2 \lambda(y-1)+3 \lambda(z-1)=0$ $\Rightarrow \quad \lambda[-2 x-2-2 y+2+3 z-3]=0$ $\Rightarrow \quad-2 x-2 y+3 z-3=0$ $\Rightarrow \quad 2 x+2 y-3 z+3=0$