The equation of the plane through given line is $ A(x-1)+B(y-2)+C(z-3)=0 $ ...(i) where A, B and C are the DRs of the normal to the plane. Since, the straight line lie on the plane. .. DRs of the plane is perpendicular to the line, ie,
$ 5A+4B+5C=0 $ ...(ii)
Since, it passes through (0, 0, 0), we get $ -A-2B-3C=0 $
$ \Rightarrow $ $ A+2B+3C=0 $ ...(iii)
On solving Eqs. (ii) and (iii), we get
$ \frac{A}{2}=\frac{B}{-10}=\frac{C}{6} $ From Eq. (i), $ 2(x-1)-10(y-2)+6(z-3)=0 $
$ \Rightarrow $ $ 2x-2-10y+20+6z-18=0 $
$ \Rightarrow $ $ 2x-10y+6z=0 $
$ \Rightarrow $ $ x-5y+3z=0 $