1. **State the problem:** Find the equation of the plane passing through points $P_1(1,2,-1)$, $P_2(2,3,1)$, and $P_3(3,-1,2)$.\n\n2. **Formula and concept:** The equation of a plane can be written as $$A(x-x_0) + B(y-y_0) + C(z-z_0) = 0$$ where $(x_0,y_0,z_0)$ is a point on the plane and $(A,B,C)$ is the normal vector to the plane.\n\n3. **Find two vectors on the plane:** Use the given points to find vectors \n$$\vec{v_1} = P_2 - P_1 = (2-1, 3-2, 1+1) = (1,1,2)$$\n$$\vec{v_2} = P_3 - P_1 = (3-1, -1-2, 2+1) = (2,-3,3)$$\n\n4. **Find the normal vector:** The normal vector is the cross product of $\vec{v_1}$ and $\vec{v_2}$:\n$$\vec{n} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 2 \\ 2 & -3 & 3 \end{vmatrix} = \mathbf{i}(1 \cdot 3 - 2 \cdot (-3)) - \mathbf{j}(1 \cdot 3 - 2 \cdot 2) + \mathbf{k}(1 \cdot (-3) - 1 \cdot 2)$$\n$$= \mathbf{i}(3 + 6) - \mathbf{j}(3 - 4) + \mathbf{k}(-3 - 2) = 9\mathbf{i} + 1\mathbf{j} - 5\mathbf{k}$$\nSo, $\vec{n} = (9,1,-5)$.\n\n5. **Write the plane equation:** Using point $P_1(1,2,-1)$ and normal vector $(9,1,-5)$, the plane equation is:\n$$9(x-1) + 1(y-2) - 5(z+1) = 0$$\nSimplify:\n$$9x - 9 + y - 2 - 5z - 5 = 0$$\n$$9x + y - 5z - 16 = 0$$\n\n**Final answer:** The equation of the plane is $$9x + y - 5z = 16$$.