The diagram shows points $A(5, 0)$, $B(a, 3a)$ and $C(-2, 6)$ such that the line $AB$ is perpendicular to the line $BC$.

1. Find the expressions for the gradients of lines $AB$ and $BC$.

Gradient of $AB$, $m_{AB} = \frac{3a - 0}{a - 5} = \frac{3a}{a - 5}$
Gradient of $BC$, $m_{BC} = \frac{3a - 6}{a - (-2)} = \frac{3a - 6}{a + 2}$

2. Apply the condition for perpendicular lines to form an equation in $a$.

Since $AB \perp BC$, $m_{AB} \times m_{BC} = -1$.
$\left(\frac{3a}{a - 5}\right) \left(\frac{3a - 6}{a + 2}\right) = -1$
$3a(3a - 6) = -1(a - 5)(a + 2)$

3. Solve the quadratic equation for $a$ and reject the invalid solution.

$9a^2 - 18a = -(a^2 - 3a - 10)$
$9a^2 - 18a = -a^2 + 3a + 10$
$10a^2 - 21a - 10 = 0$
$(2a - 5)(5a + 2) = 0$
$a = 2.5$ or $a = -0.4$
From the diagram, point $B(a, 3a)$ is in the first quadrant, so $a > 0$.
Therefore, $a = 2.5$ (shown).

1. Find the coordinates of the midpoint of $AC$

Midpoint of $AC = \left(\frac{5 + (-2)}{2}, \frac{0 + 6}{2}\right) = (1.5, 3)$

2. Set up equations for the coordinates of $D$ using the property that diagonals of a parallelogram bisect each other

In a parallelogram $ABCD$, the midpoint of diagonal $BD$ is the same as the midpoint of diagonal $AC$.
From part (a), $a = 2.5$, so $B$ is $(2.5, 3(2.5)) = (2.5, 7.5)$.
Let the coordinates of $D$ be $(x, y)$.
$\frac{x + 2.5}{2} = 1.5$
$\frac{y + 7.5}{2} = 3$

3. Solve for the coordinates of $D$

$x + 2.5 = 3 \implies x = 0.5$
$y + 7.5 = 6 \implies y = -1.5$
The coordinates of $D$ are $(0.5, -1.5)$.

1. Set up the equation for the area of triangle $AEC$ using the shoelace formula.

Using the coordinates $A(5, 0)$, $C(-2, 6)$, and $E(x, 2x)$:
$\text{Area} = \frac{1}{2} \left| \begin{matrix} 5 & -2 & x & 5 \\ 0 & 6 & 2x & 0 \end{matrix} \right|$
$\text{Area} = \frac{1}{2} |(5)(6) + (-2)(2x) + (x)(0) - (0)(-2) - (6)(x) - (2x)(5)|$
$\text{Area} = \frac{1}{2} |30 - 4x - 6x - 10x|$
$\text{Area} = \frac{1}{2} |30 - 20x|$
$\text{Area} = |15 - 10x|$

2. Equate the area to 5 and solve for $x$.

Given that the area is 5 units$^2$:
$|15 - 10x| = 5$
This gives two possible equations:
$15 - 10x = 5 \quad \text{or} \quad 15 - 10x = -5$
$10x = 10 \quad \text{or} \quad 10x = 20$
$x = 1 \quad \text{or} \quad x = 2$

3. Select the valid value of $x$ based on the given condition.

Since it is given that $x > 1$, we reject $x = 1$.
Therefore, $x = 2$.