A plane passes through $(2,3,-1)$ and is perpendicular to the line having direction ratios $3,-4,7$. The perpendicular distance from the origin to this plane is

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 $\begin{aligned} & =\frac{|3 \times 0-4 \times 0+7 \times 0+13|}{\sqrt{9+16+49}} \\ & =\frac{13}{\sqrt{74}} \end{aligned}$