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

$x+3y + 6z = 7 $

The equation of the plane containing the line $2x - 5y + z - 3= 0 = x + y + 4z - 5 $ is

$(2x -5y + z - 3) + λ (x + y + 4z -5) = 0 $

or, $ x (λ + 2) + y ( λ - 5) + z (4λ + 1) - ( 5λ + 3) = 0 $ ......(i)

This parallel to the plane $x + 3y + 6z - 1 = 0 .$

$∴ \frac{λ+2}{1}=\frac{λ-5}{3}=\frac{4λ+1}{6}⇒ λ = - \frac{11}{2}$

Putting $ λ = -\frac{11}{2}$ in (i), we obtain $ x + 3y + 6z = 7 $ as the equation of the required plane.