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

1. Problem Description

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

2. Solution Steps

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

First, we solve the system of equations formed by the two plane equations:

$x - 2y = 0$

$2x + 3y - 2 = 0$

From the first equation, we have $x = 2y$.

Substitute this into the second equation:

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

$4y + 3y - 2 = 0$

$7y = 2$

$y = \frac{2}{7}$

Now, substitute $y = \frac{2}{7}$ back into $x = 2y$:

$x = 2(\frac{2}{7}) = \frac{4}{7}$

So, a point on the line is $(\frac{4}{7}, \frac{2}{7}, 0)$.

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

The normal vector of plane (a) is $\vec{n_1} = (1, -2, 0)$.

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

The direction vector $\vec{d}$ is $\vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 0 \\ 2 & 3 & -2 \end{vmatrix} = (4, 2, 7)$.

The equation of the line can be expressed in parametric form as:

$x = \frac{4}{7} + 4t$

$y = \frac{2}{7} + 2t$

$z = 0 + 7t$

3. Final Answer

The parametric equation of the line is:

$x = \frac{4}{7} + 4t$

$y = \frac{2}{7} + 2t$

$z = 7t$