1. **Problem Statement:** Find the canonical equations of the straight line defined by the system of planes: $$\begin{cases} 3x + y + z - 2 = 0 \\ 2x - y - 3z + 6 = 0 \end{cases}$$ 2. **Concept:** The canonical form of a line in 3D is given by: $$\frac{x - x_0}{l} = \frac{y - y_0}{m} = \frac{z - z_0}{n}$$ where $(x_0,y_0,z_0)$ is a point on the line and $(l,m,n)$ is the direction vector. 3. **Step 1: Find the direction vector** The direction vector is perpendicular to the normal vectors of both planes. Normals: $$\vec{n_1} = (3,1,1), \quad \vec{n_2} = (2,-1,-3)$$ Direction vector $\vec{d} = \vec{n_1} \times \vec{n_2}$: $$\vec{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 1 & 1 \\ 2 & -1 & -3 \end{vmatrix} = (1 \cdot (-3) - 1 \cdot (-1), -(3 \cdot (-3) - 1 \cdot 2), 3 \cdot (-1) - 1 \cdot 2)$$ $$= (-3 + 1, -( -9 - 2), -3 - 2) = (-2, 11, -5)$$ 4. **Step 2: Find a point on the line** Solve the system for a point $(x_0,y_0,z_0)$: From the first equation: $$3x + y + z = 2$$ From the second: $$2x - y - 3z = -6$$ Add both equations: $$3x + y + z + 2x - y - 3z = 2 - 6 \Rightarrow 5x - 2z = -4$$ Express $x$: $$x = \frac{-4 + 2z}{5}$$ Substitute into first equation: $$3\left(\frac{-4 + 2z}{5}\right) + y + z = 2 \Rightarrow \frac{-12 + 6z}{5} + y + z = 2$$ Multiply both sides by 5: $$-12 + 6z + 5y + 5z = 10 \Rightarrow 5y + 11z = 22$$ Express $y$: $$y = \frac{22 - 11z}{5}$$ Choose $z=0$ for simplicity: $$x_0 = \frac{-4}{5}, \quad y_0 = \frac{22}{5}, \quad z_0 = 0$$ 5. **Step 3: Write canonical equations** $$\frac{x - (-\frac{4}{5})}{-2} = \frac{y - \frac{22}{5}}{11} = \frac{z - 0}{-5}$$ Or equivalently: $$\frac{x + \frac{4}{5}}{-2} = \frac{y - \frac{22}{5}}{11} = \frac{z}{-5}$$ **Final answer:** $$\boxed{\frac{x + \frac{4}{5}}{-2} = \frac{y - \frac{22}{5}}{11} = \frac{z}{-5}}$$