Given:

$x+3y+2z=9$

$\Rightarrow \frac{x}{9}+\frac{y}{3}+\frac{z}{\left({\displaystyle \frac{9}{2}}\right)}=1$

Comparing with the standard form $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$, we get

$a=9\to \left(x-\text{intercept}\right)$

$b=3\to \left(y-\text{intercept}\right)$

$c=\left(\frac{9}{2}\right)\to \left(z-\text{intercept}\right)$

So, the normal vector for the required plane is $\overrightarrow{n}=9\hat{i}+3\hat{j}+\frac{9}{2}\hat{k}$

$\overrightarrow{a}=3\hat{i}+5\hat{j}+7\hat{k}$

Equation of required plane,

$\Rightarrow \overrightarrow{r}·\overrightarrow{n}=\overrightarrow{a}·\overrightarrow{n}$

$\Rightarrow \overrightarrow{r}·\left(9\hat{i}+3\hat{j}+\frac{9}{2}\hat{k}\right)=\left(3\hat{i}+5\hat{j}+7\hat{k}\right)·\left(9\hat{i}+3\hat{j}+\frac{9}{2}\hat{k}\right)$

$\Rightarrow \overrightarrow{r}·\left(9\hat{i}+3\hat{j}+\frac{9}{2}\hat{k}\right)=27+15+\frac{63}{2}$

$\Rightarrow \left(x\hat{i}+y\hat{j}+z\hat{k}\right)·\left(9\hat{i}+3\hat{j}+\frac{9}{2}\hat{k}\right)=42+\frac{63}{2}$

$\Rightarrow 9x+3y+\frac{9}{2}z=\frac{147}{2}$

$\Rightarrow 3x+y+\frac{3}{2}z=\frac{49}{2}$

$\Rightarrow 6x+2y+3z=49$.