The required plane contains the line of intersection of \(2x-5y+z=3\) and \(x+y+4z=5\), so it can be written as \((2x-5y+z-3) + \lambda(x+y+4z-5) = 0\), which is \((2+\lambda)x + (-5+\lambda)y + (1+4\lambda)z - (3+5\lambda) = 0\). For this to be parallel to \(x+3y+6z=1\), the normal vectors must be proportional: \(\frac{2+\lambda}{1} = \frac{-5+\lambda}{3} = \frac{1+4\lambda}{6}\). From the first two: \(3(2+\lambda) = -5+\lambda\), giving \(6+3\lambda = -5+\lambda\), so \(2\lambda = -11\) and \(\lambda = -\frac{11}{2}\). Substituting back: the plane is \((2-\frac{11}{2})x + (-5-\frac{11}{2})y + (1-22)z - (3-\frac{55}{2}) = 0\), which simplifies to \(-\frac{7}{2}x - \frac{21}{2}y - 21z + \frac{49}{2} = 0\). Multiplying by \(-\frac{2}{7}\): \(x + 3y + 6z = 7\).