The problem asks us to find the equation of the line formed by the intersection of two planes: plane $(a): x - 2y - z = 0$ and plane $(\beta): 2x + 3y - z = 0$.

1. Problem Description

The problem asks us to find the equation of the line formed by the intersection of two planes: plane $(a): x - 2y - z = 0$ and plane $(\beta): 2x + 3y - z = 0$.

2. Solution Steps

To find the equation of the line of intersection, we need to find the direction vector and a point on the line.

First, let's find the direction vector of the line of intersection. This vector is perpendicular to the normal vectors of both planes. The normal vector of plane $(a)$ is $\vec{n_1} = <1, -2, -1>$, and the normal vector of plane $(\beta)$ is $\vec{n_2} = <2, 3, -1>$.

The direction vector $\vec{d}$ of the line of intersection can be found by taking the cross product of the normal vectors:

$\vec{d} = \vec{n_1} \times \vec{n_2} = <1, -2, -1> \times <2, 3, -1>$

$\vec{d} = <(-2)(-1) - (-1)(3), (-1)(2) - (1)(-1), (1)(3) - (-2)(2)>$

$\vec{d} = <2 + 3, -2 + 1, 3 + 4>$

$\vec{d} = <5, -1, 7>$

Now, we need to find a point on the line of intersection. We can do this by solving the system of equations formed by the plane equations:

$x - 2y - z = 0$

$2x + 3y - z = 0$

Subtracting the first equation from the second gives:

$(2x + 3y - z) - (x - 2y - z) = 0 - 0$

$x + 5y = 0$

$x = -5y$

Substitute $x = -5y$ into the first equation:

$(-5y) - 2y - z = 0$

$-7y - z = 0$

$z = -7y$

Let's choose $y = 0$. Then $x = -5(0) = 0$ and $z = -7(0) = 0$. So, the point $(0, 0, 0)$ lies on the line.

Thus, the line of intersection passes through the origin $(0, 0, 0)$ and has direction vector $<5, -1, 7>$. The parametric equations of the line are:

$x = 5t$

$y = -t$

$z = 7t$

3. Final Answer

The parametric equations of the line of intersection are $x = 5t, y = -t, z = 7t$.