Find the Cartesian equation of the plane which passes through the point $(5,2,-4)$ and is perpendicular to the line with direction ratios $2,3,-1$?

The direction ratios of the line perpendicular to the plane are the direction ratios of the normal to the plane.So, $(a,b,c)=(2,3,-1)$.The equation of a plane passing through $(x_1,y_1,z_1)$ with normal $(a,b,c)$ is:
$$a(x-x_1)+b(y-y_1)+c(z-z_1)=0$$
Substitute the values:
$$2(x-5)+3(y-2)+(-1)(z-(-4))=0$$
$$2x-10+3y-6-(z+4)=0$$
$$2x+3y-16-z-4=0$$
$$2x+3y-z=20$$