Question 19: Two lines L1: 2y − 3x − 6 = 0 and L2: 3y + x = 20 intersect at a point M.

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

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

Substitute this expression for $x$ into Equation 1:

Substitute the value of $y$ back into the expression for $x$:

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

Rewrite $L_2: 3y + x = 20$ in the form $y = mx + c$:

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

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

$L_3$ passes through point M $(2, 6)$ and has a gradient of 3. Using the point-slope form $y - y_1 = m(x - x_1)$:

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

Rewrite $L_1: 2y - 3x - 6 = 0$ in the form $y = mx + c$:

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

Since $L_4$ is parallel to $L_1$, their gradients are equal.
The gradient of $L_4$ ($m_4$) is $\frac{3}{2}$.

$L_4$ passes through $(-1, 3)$ and has a gradient of $\frac{3}{2}$. Using the point-slope form $y - y_1 = m(x - x_1)$:

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

The y-intercept is $\boxed{\left(0, \frac{9}{2}\right)}$ or $\boxed{(0, 4.5)}$.

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

Multiply both sides by 2:

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