The equation of the plane passes through the point $(2,3,-1)$ is


where $a, b, c$ are the direction ratio of the normal to the plane.

Also, given the plane is perpendicular to the line whose direction ratio is $(3,-4,7)$. So, that line and the normal of the plane are parallel.
$\begin{array}{lc}
\Rightarrow & \frac{a}{3}=\frac{b}{-4}=\frac{c}{7}=k \\
\Rightarrow & a=3 k, b=-4 k, c=7 k
\end{array}$
From Eq. (i)
$3 k(x-2)-4 k(y-3)+7 k(z+1)=0$

Now, perpendicular distance from the origin to this plane
$\begin{aligned}
& =\frac{|3 \times 0-4 \times 0+7 \times 0+13|}{\sqrt{9+16+49}} \\
& =\frac{13}{\sqrt{74}}
\end{aligned}$