1. **State the problem:** Find the equations of the line of intersection of the planes given by $$x + 2y + 3z = 6$$ and $$x - y - z = 4$$. 2. **Recall the formula and rules:** The line of intersection of two planes can be found by solving the system of equations simultaneously. The direction vector of the line is perpendicular to the normal vectors of both planes, so it is the cross product of the normals. 3. **Find the normal vectors:** - For plane 1: normal vector $\mathbf{n_1} = (1, 2, 3)$ - For plane 2: normal vector $\mathbf{n_2} = (1, -1, -1)$ 4. **Find the direction vector $\mathbf{d}$ of the line:** $$\mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 1 & -1 & -1 \end{vmatrix}$$ Calculate the determinant: $$\mathbf{d} = \mathbf{i}(2 \times -1 - 3 \times -1) - \mathbf{j}(1 \times -1 - 3 \times 1) + \mathbf{k}(1 \times -1 - 2 \times 1)$$ $$= \mathbf{i}(-2 + 3) - \mathbf{j}(-1 - 3) + \mathbf{k}(-1 - 2)$$ $$= \mathbf{i}(1) - \mathbf{j}(-4) + \mathbf{k}(-3) = (1, 4, -3)$$ 5. **Find a point on the line:** Solve the system for a specific value to find a point. Set $z = 0$: From plane 1: $$x + 2y = 6$$ From plane 2: $$x - y = 4$$ Solve the system: From plane 2: $$x = 4 + y$$ Substitute into plane 1: $$4 + y + 2y = 6$$ $$3y + 4 = 6$$ $$3y = 2$$ $$y = \frac{2}{3}$$ Then: $$x = 4 + \frac{2}{3} = \frac{14}{3}$$ So the point is $\left(\frac{14}{3}, \frac{2}{3}, 0\right)$. 6. **Write parametric equations of the line:** Using point $P_0 = \left(\frac{14}{3}, \frac{2}{3}, 0\right)$ and direction vector $\mathbf{d} = (1, 4, -3)$, the parametric equations are: $$x = \frac{14}{3} + t$$ $$y = \frac{2}{3} + 4t$$ $$z = 0 - 3t = -3t$$ **Final answer:** The line of intersection is given by $$\boxed{\begin{cases} x = \frac{14}{3} + t \\ y = \frac{2}{3} + 4t \\ z = -3t \end{cases}}$$