If the equation of the plane passing through the point $(2,-1,3)$ and perpendicular to the planes $3 x-2 y+z=9$ and $x+y+z=9$ is $x+b y+c z+d=0$, then $d=$

Required plane passes through $(2,-1,3)$. Let $a, b$ and $c$ are Dr's of normal to required plane. Equation of required plane $$ a(x-2)+b(y+1)+c(z-3)=0 $$ Since, plane $\perp$ to $3 x-2 y+z=9$ $\therefore \quad 3 a-2 b+c=0$ $$ \left[\because a_1 a_2+b_1 b_2+c_1 c_2=0\right] $$ Now, required plane is also $\perp$ to $x+y+z=9$ $\therefore$ $$ a+b+c=0 $$ From Eqs. (ii) and (iii), we get $$ \begin{array}{rlrl} & & \frac{a}{-2-1}=\frac{b}{1-3}=\frac{c}{3+2} \\ \Rightarrow \quad & \frac{a}{-3}=\frac{b}{-2}=\frac{c}{5}=\lambda \end{array} $$ So, $a=-3 \lambda, b=-2 \lambda$ and $c=5 \lambda$ Put in Eq. (i), $$ \begin{array}{rlrl} & -3 \lambda(x-2)-2 \lambda(y+1)+5 \lambda(z-3) & =0 \\ \Rightarrow & \lambda(-3(x-2)-2(y+1)+5(z-3)] & =0 \\ \Rightarrow & & -3 x+6-2 y-2+5 z-15 & =0 \\ \Rightarrow & & -3 x-2 y+5 z-11 & =0 \\ \Rightarrow & & x+\frac{2}{3} y-\frac{5}{3} z+\frac{11}{3} & =0 \end{array} $$ By comparing $a x+b y+c z+d=0$, we get $$ d=11 / 3 $$