First, solve the system of equations:

1. \(y^2 = 12x\)

2. \(3y = 4x + 6\)

From equation (2), express \(x\) in terms of \(y\):

\(x = \frac{3y - 6}{4}\)

Substitute \(x\) in equation (1):

\(y^2 = 12 \left( \frac{3y - 6}{4} \right)\)

Simplify:

\(y^2 = 9y - 18\)

Rearrange to form a quadratic equation:

\(y^2 - 9y + 18 = 0\)

Factor the quadratic:

\((y - 3)(y - 6) = 0\)

Thus, \(y = 3\) or \(y = 6\).

Find corresponding \(x\) values:

For \(y = 3\):

\(x = \frac{3(3) - 6}{4} = \frac{9 - 6}{4} = \frac{3}{4}\)

For \(y = 6\):

\(x = \frac{3(6) - 6}{4} = \frac{18 - 6}{4} = 3\)

The points of intersection are \(\left( \frac{3}{4}, 3 \right)\) and \((3, 6)\).

Calculate the distance between the points:

\(\text{Distance} = \sqrt{\left( 3 - \frac{3}{4} \right)^2 + (6 - 3)^2}\)

\(= \sqrt{\left( \frac{9}{4} \right)^2 + 3^2}\)

\(= \sqrt{\frac{81}{16} + 9}\)

\(= \sqrt{\frac{81}{16} + \frac{144}{16}}\)

\(= \sqrt{\frac{225}{16}}\)

\(= \frac{15}{4}\)

\(= 3.75\)