Let the plane $\pi$ pass through the point $(1,0,1)$ and perpendicular to the planes $2 x+3 y-z=2$ and $x-y+2 z=1$. Let the equation of the plane passing through the point $(11,7,5)$ and parallel to the plane $\pi$ be $a x+b y-z-d=0$. Then, $\frac{a}{b}+\frac{b}{d}=$

The equation of plane passes through the point $(1,0,1)$ is $a(x-1)+b(y-0)+c(z-1)=0$ ...(i) If this plane is perpendicular to the planes $2 x+3 y-z=2$ and $x-y+2 z=1$ Then, $2 a+3 b-c=2$ and $a-b+2 c=1$ $\begin{aligned} & \Rightarrow \frac{a}{6-1}=\frac{b}{-1-4}=\frac{c}{2(-1)-3} \\ & \Rightarrow \frac{a}{5}=\frac{b}{-3}=\frac{c}{-5}=k\end{aligned}$ Let $a=5 k, b=-3 k, c=-5 k$ Putting the values of $a, b$ and $c$ in Eq. (i), we get the required equation of plane $5 k(x-1)+(-3 k)(y)+(-5 k)(z-1)=0$ $\begin{aligned} & \Rightarrow \quad 5 x-5-3 y-5 z+5=0 \\ & \Rightarrow \quad 5 x-3 y-5 z=0\end{aligned}$ Now, plane passing through $(11,7,5)$ and parallel to the plane $\pi(5 x-3 y-5 z)$. Equation of a plane parallel to $5 x-3 y-5 z=k$ ...(ii) and passing through $(11,7,5)$ $5(11)-3(7)-5(5)=k$ $\Rightarrow \quad 55-21-25=k$ $\Rightarrow \quad k=9$ So, equation of plane $\Rightarrow \quad 5 x-3 y-5 z=9$ Comparing with $a x+b y-z+d=0$ (given) So, $a=5, b=-3, c=5, d=9$ $\frac{a}{b}+\frac{b}{d}=\frac{5}{-3}+\frac{-3}{9}$ $\Rightarrow \frac{a}{b}+\frac{b}{d}=\frac{-15-3}{9}=\frac{-18}{9}=-2$