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
$$
d=11 / 3
$$