Find the coordinates of M where lines L1 and L2 intersect, and the equation of line L3 perpendicular to L2 at M.

Line $L_1: 2y - 3x - 6 = 0 \implies 2y - 3x = 6$ (Equation 1)
Line $L_2: 3y + x = 20$ (Equation 2)

From Equation 2, express $x$ in terms of $y$:

Substitute this expression for $x$ into Equation 1:

Now substitute $y=6$ back into the expression for $x$:

The coordinates of M are $\boxed{(2, 6)}$.

The equation of $L_2$ is $3y + x = 20$.
Rearrange to $y = mx + c$ form:

The gradient of $L_2$ is $m_2 = -\frac{1}{3}$.

Since $L_3$ is perpendicular to $L_2$, the product of their gradients is $-1$:

Using the point-slope form $y - y_1 = m(x - x_1)$:

The equation of $L_3$ is $\boxed{y = 3x}$.

The equation of $L_1$ is $2y - 3x - 6 = 0$.
Rearrange to $y = mx + c$ form:

The gradient of $L_1$ is $m_1 = \frac{3}{2}$.

Since $L_4$ is parallel to $L_1$, their gradients are equal:

Using the point-slope form $y - y_1 = m(x - x_1)$:

Multiply the entire equation by 2 to eliminate the fraction:

For the y-intercept, set $x = 0$:

The y-intercept is $\boxed{(0, 4.5)}$.

For the x-intercept, set $y = 0$:

Multiply the entire equation by 2:

The x-intercept is $\boxed{(-3, 0)}$.