Step 1:

The given points are $P(3,4,1)$ and $Q(0,1,0)$

Equation of the plane passing through $(3,4,1)$ is

$a(x-3)+b(y-4)+c(2-1)=0$-------(1)

This plane passes through $(0,1,0)$

$\therefore a(0-3)+b(1-4)+c(0-1)=0$

$-3a-3b-c=0$

(i.e) $3a+3b+c=0$-----(2)

Step 2:

The plane (equ(1)) is parallel to the given line

$\large\frac{x+3}{2}=\frac{y-3}{7}=\frac{z-2}{5}$

$\therefore$ Normal to the plane is perpendicular to the line.

$\therefore a(2)+7(b)+c(5)=0$

$\Rightarrow 2a+7b+5c=0$------(3)

Step 3:

Now let us solve equ(2) and equ(3)

$\large\frac{a}{\begin{vmatrix}3 & 1\\7 & 5\end{vmatrix}}=\large\frac{b}{\begin{vmatrix}1 & 3\\5& 2\end{vmatrix}}=\large\frac{c}{\begin{vmatrix}3 & 3\\2 & 7\end{vmatrix}}$

(i.e) $\large\frac{a}{8}=\frac{b}{-13}=\frac{c}{15}$$=k$

$a=8k$

$b=-13k$

$c=15k$

Step 4:

Substituting these values in equ(1) we get,

$8(x-3)-13(y-4)+15(2-1)=0$

$\Rightarrow 8x-13y+15z+12=0$

This is the required equation of the plane.