1. **State the problem:** We need to find the equation of a plane that passes through the point $P(2,1,-3)$ and contains the vectors $\vec{v_1} = 3\mathbf{i} + \mathbf{j} + 4\mathbf{k}$ and $\vec{v_2} = 4\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}$. 2. **Formula and important rules:** 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. 3. **Find the normal vector:** The normal vector $\mathbf{n}$ is perpendicular to both $\vec{v_1}$ and $\vec{v_2}$, so it is their cross product: $$\mathbf{n} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 1 & 4 \\ 4 & 2 & 3 \end{vmatrix}$$ 4. **Calculate the determinant:** $$\mathbf{n} = \mathbf{i}(1 \times 3 - 4 \times 2) - \mathbf{j}(3 \times 3 - 4 \times 4) + \mathbf{k}(3 \times 2 - 1 \times 4)$$ $$= \mathbf{i}(3 - 8) - \mathbf{j}(9 - 16) + \mathbf{k}(6 - 4)$$ $$= -5\mathbf{i} + 7\mathbf{j} + 2\mathbf{k}$$ 5. **Write the plane equation:** Using point $P(2,1,-3)$ and normal vector $\mathbf{n} = (-5, 7, 2)$: $$-5(x - 2) + 7(y - 1) + 2(z + 3) = 0$$ 6. **Simplify:** $$-5x + 10 + 7y - 7 + 2z + 6 = 0$$ $$-5x + 7y + 2z + (10 - 7 + 6) = 0$$ $$-5x + 7y + 2z + 9 = 0$$ **Final answer:** $$\boxed{-5x + 7y + 2z + 9 = 0}$$