To find the parametric equations for the line of intersection of the two planes, we will first follow these steps:

1. Step 1: Represent the planes as vector equations:

The equations of the planes are:

Plane 1: $2x - 3y + z = 6$
Plane 2: $x - 2y + z = 5$

We can represent these equations as:

$\mathbf{n}_1 = \langle 2, -3, 1 \rangle \quad \text{(normal vector of the first plane)}$ $\mathbf{n}_2 = \langle 1, -2, 1 \rangle \quad \text{(normal vector of the second plane)}$

2. Step 2: Find the direction vector of the line of intersection:

The direction vector of the line of intersection is given by the cross product of the normal vectors of the two planes.

$\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2$

Now, we compute the cross product:

$\mathbf{d} = \langle 2, -3, 1 \rangle \times \langle 1, -2, 1 \rangle$

We use the determinant of a 3x3 matrix to compute the cross product:

2 & -3 & 1 \\ 1 & -2 & 1 \end{vmatrix}$$ Expanding the determinant: $$\mathbf{d} = \mathbf{i} \left( (-3)(1) - (1)(-2) \right) - \mathbf{j} \left( (2)(1) - (1)(1) \right) + \mathbf{k} \left( (2)(-2) - (-3)(1) \right)$$ $$\mathbf{d} = \mathbf{i} \left( -3 + 2 \right) - \mathbf{j} \left( 2 - 1 \right) + \mathbf{k} \left( -4 + 3 \right)$$ $$\mathbf{d} = \mathbf{i} (-1) - \mathbf{j} (1) + \mathbf{k} (-1)$$ $$\mathbf{d} = \langle -1, -1, -1 \rangle$$ Thus, the direction vector of the line of intersection is $$\mathbf{d} = \langle -1, -1, -1 \rangle$$. 3. **Step 3: Find a point on the line of intersection:** To find a point on the line, we need to solve the system of equations for the two planes. We can do this by setting one of the variables as a parameter, say $$z = t$$, and solving for $$x$$ and $$y$$ in terms of $$t$$. From Plane 1: $$2x - 3y + z = 6 \quad \Rightarrow \quad 2x - 3y + t = 6 \quad \Rightarrow \quad 2x - 3y = 6 - t \quad (1)$$ From Plane 2: $$x - 2y + z = 5 \quad \Rightarrow \quad x - 2y + t = 5 \quad \Rightarrow \quad x - 2y = 5 - t \quad (2)$$ Now, we solve the system of equations (1) and (2) for $$x$$ and $$y$$: Multiply equation (2) by 2: $$2x - 4y = 10 - 2t \quad (3)$$ Subtract equation (1) from equation (3): $$(2x - 4y) - (2x - 3y) = (10 - 2t) - (6 - t)$$ $$- y = 4 - t$$ $$y = t - 4$$ Substitute $$y = t - 4$$ into equation (2): $$x - 2(t - 4) = 5 - t$$ $$x - 2t + 8 = 5 - t$$ $$x = 5 - t + 2t - 8$$ $$x = t - 3$$ Thus, the parametric equations for the line of intersection are: $$x = t - 3, \quad y = t - 4, \quad z = t$$ 4. **Final Parametric Equations:** The parametric equations for the line of intersection of the two planes are: $$x = t - 3, \quad y = t - 4, \quad z = t$$ --- Would you like any further clarifications or details on the process? Here are five related questions for deeper understanding: 1. How do we interpret the direction vector in the context of the line of intersection? 2. Can we find parametric equations for the line if given three planes instead of two? 3. How would the solution change if the two planes were parallel or coincident? 4. What other methods can be used to find the line of intersection of two planes? 5. How does the choice of parameter $$z = t$$ affect the solution? **Tip**: When solving a system of equations, setting one variable as a parameter is a common technique in finding parametric forms.