1. **State the problem:** Find the equation of the plane passing through the point $(1,-1,3)$ and containing the line given by $$\frac{x}{3} = \frac{y+2}{2} = \frac{z-1}{4}.$$\n\n2. **Understand the line:** The line can be parameterized as $$x=3t,\quad y=-2+2t,\quad z=1+4t,$$ where $t$ is a parameter.\n\n3. **Identify points and vectors:** The line passes through point $P_0=(0,-2,1)$ when $t=0$. The direction vector of the line is $$\vec{d} = (3,2,4).$$\n\n4. **Find vector from $P_0$ to given point $P=(1,-1,3)$:** $$\vec{v} = P - P_0 = (1-0, -1+2, 3-1) = (1,1,2).$$\n\n5. **The plane contains the line and point $P$, so its normal vector $\vec{n}$ is perpendicular to both $\vec{d}$ and $\vec{v}$. Find $\vec{n}$ by cross product:** $$\vec{n} = \vec{d} \times \vec{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 2 & 4 \\ 1 & 1 & 2 \end{vmatrix} = (2\cdot 2 - 4\cdot 1, 4\cdot 1 - 3\cdot 2, 3\cdot 1 - 2\cdot 1) = (4-4, 4-6, 3-2) = (0, -2, 1).$$\n\n6. **Equation of the plane:** Using point $P_0=(0,-2,1)$ and normal vector $\vec{n}=(0,-2,1)$, the plane equation is $$0(x-0) - 2(y+2) + 1(z-1) = 0,$$ which simplifies to $$-2y -4 + z -1 = 0,$$ or $$-2y + z -5 = 0.$$\n\n7. **Final answer:** The equation of the plane is $$\boxed{-2y + z = 5}.$$