The equation of the plane containing the line $$2x-5y+z=3; \,x+y+4z=5,$$       and parallel to the plane, $$x+3y+6z=1,$$    is :

Solution :
Equation of the plane containing the lines
$$\eqalign{ & 2x - 5y + z = 3{\text{ and }}x + y + 4z = 5{\text{ is}} \cr & 2x - 5y + z - 3 + \lambda \left( {x + y + 4z - 5} \right){\text{ = 0}} \cr & \Rightarrow \left( {2 + \lambda } \right)x + \left( { - 5 + \lambda } \right)y + \left( {1 + 4\lambda } \right)z + \left( { - 3 - 5\lambda } \right) = 0.....({\text{i}}) \cr} $$
Since the plane (i) parallel to the given plane $$x+3y+6z=1$$
$$\therefore \frac{{2 + \lambda }}{1} = \frac{{ - 5 + \lambda }}{3} = \frac{{1 + 4\lambda }}{6}\,\,\,\,\, \Rightarrow \lambda = - \frac{{11}}{2}$$
Hence equation of the required plane is
$$\eqalign{ & \left( {2 - \frac{{11}}{2}} \right)x + \left( { - 5 - \frac{{11}}{2}} \right)y + \left( {1 - \frac{{44}}{2}} \right)z + \left( { - 3 + \frac{{55}}{2}} \right) = 0 \cr & \Rightarrow x + 3y + 6z = 7 \cr} $$