Find the equation of the plane passing through the point (–2, 1, 3) and having (3, –5, 4) as direction ratios of its normal.

d.r’s of normal to the plane (a, b, c)= (3, –5, 4)
Given  passing through the point (x₁, y₁, z₁) = (-2,1,3)
Equation of  the plane passing through (x₁, y₁, z₁)and having d.r’s (a,b,c) is
$\begin{array}{l}\mathrm{a}\left(\mathrm{x}-{\mathrm{x}}_1\right)+\mathrm{b}\left(\mathrm{y}-{\mathrm{y}}_1\right)+\mathrm{c}\left(\mathrm{z}-{\mathrm{z}}_1\right)=0\\ \Rightarrow 3(\mathrm{x}+2)-5(\mathrm{y}-1)+4(\mathrm{z}-3)=0\\ \Rightarrow 3\mathrm{x}+6-5\mathrm{y}+5+4\mathrm{z}-12=0\\ \Rightarrow 3\mathrm{x}-5\mathrm{y}+4\mathrm{z}-1=0\end{array}$