The problem is to find the equation of the line of intersection between two planes. The equations of the planes are given as: Plane $(\alpha)$: $x - 2y - z = 0$ Plane $(\beta)$: $2x + 3y - 2z = 0$

1. Problem Description

The problem is to find the equation of the line of intersection between two planes. The equations of the planes are given as:

Plane $(\alpha)$: $x - 2y - z = 0$

Plane $(\beta)$: $2x + 3y - 2z = 0$

2. Solution Steps

The line of intersection can be found by finding two points on the line. Since both planes pass through the origin $(0,0,0)$, this is one point on the line of intersection. We need to find another point. To do this, we can set one of the variables, say $z$, to a non-zero value and solve for $x$ and $y$.

Let $z = 1$. Then the equations become:

$x - 2y - 1 = 0$

$2x + 3y - 2 = 0$

We can rewrite these as:

$x - 2y = 1$

$2x + 3y = 2$

Multiply the first equation by -2:

$-2x + 4y = -2$

$2x + 3y = 2$

Add the two equations:

$7y = 0$

$y = 0$

Substitute $y = 0$ into $x - 2y = 1$:

$x - 2(0) = 1$

$x = 1$

So, the second point on the line is $(1, 0, 1)$.

Now we have two points on the line: $(0, 0, 0)$ and $(1, 0, 1)$.

The direction vector of the line is the vector connecting these two points:

$\vec{v} = (1 - 0, 0 - 0, 1 - 0) = (1, 0, 1)$

The equation of the line can be written as:

$\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$

where $(x_0, y_0, z_0)$ is a point on the line and $(a, b, c)$ is the direction vector.

Using the point $(0, 0, 0)$ and the direction vector $(1, 0, 1)$, we get:

$\frac{x - 0}{1} = \frac{y - 0}{0} = \frac{z - 0}{1}$

$\frac{x}{1} = \frac{y}{0} = \frac{z}{1}$

$x = z$, and $y = 0$

Alternatively, we can find the cross product of the normal vectors of the two planes.

The normal vector of plane $(\alpha)$ is $\vec{n_1} = (1, -2, -1)$

The normal vector of plane $(\beta)$ is $\vec{n_2} = (2, 3, -2)$

$\vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & -1 \\ 2 & 3 & -2 \end{vmatrix} = \hat{i}((-2)(-2) - (-1)(3)) - \hat{j}((1)(-2) - (-1)(2)) + \hat{k}((1)(3) - (-2)(2))$

$= \hat{i}(4 + 3) - \hat{j}(-2 + 2) + \hat{k}(3 + 4) = 7\hat{i} - 0\hat{j} + 7\hat{k} = (7, 0, 7)$

The direction vector of the line is $(7, 0, 7)$, which is proportional to $(1, 0, 1)$.

Thus the line can be represented as:

$\frac{x}{1} = \frac{y}{0} = \frac{z}{1}$

So $x = z$ and $y = 0$.

3. Final Answer

$x = z$, $y = 0$