If the points $A(1, -2)$,$B(2, 3)$,$C(a, 2)$,and $D(-4, -3)$ form a parallelogram,find the value of $a$ and the height of the parallelogram taking $AB$ as the base.

(A) In a parallelogram,the diagonals bisect each other,which means the mid-point of $AC$ is equal to the mid-point of $BD$.
$\Rightarrow \left(\frac{1+a}{2}, \frac{-2+2}{2}\right) = \left(\frac{2-4}{2}, \frac{3-3}{2}\right)$
$\Rightarrow \frac{1+a}{2} = \frac{-2}{2} = -1$
$1+a = -2 \Rightarrow a = -3$
So,the value of $a$ is $-3$.
Given $AB$ as the base,let $DP$ be the height where $P$ is the foot of the perpendicular from $D$ to $AB$.
The equation of line $AB$ passing through $(1, -2)$ and $(2, 3)$ is:
$(y - (-2)) = \frac{3 - (-2)}{2 - 1}(x - 1)$
$y + 2 = 5(x - 1) \Rightarrow 5x - y = 7$ ... $(i)$
The slope of $AB$ is $m_1 = 5$. Since $DP \perp AB$,the slope of $DP$ is $m_2 = -1/5$.
The equation of line $DP$ passing through $D(-4, -3)$ with slope $-1/5$ is:
$(y + 3) = -\frac{1}{5}(x + 4)$
$5y + 15 = -x - 4 \Rightarrow x + 5y = -19$ ... $(ii)$
Solving $(i)$ and $(ii)$ for intersection point $P$:
$x + 5(5x - 7) = -19 \Rightarrow 26x = 16 \Rightarrow x = 8/13$
$y = 5(8/13) - 7 = -51/13$
$P = (8/13, -51/13)$. The height $DP$ is the distance between $D(-4, -3)$ and $P(8/13, -51/13)$:
$DP = \sqrt{(\frac{8}{13} + 4)^2 + (-\frac{51}{13} + 3)^2} = \sqrt{(\frac{60}{13})^2 + (-\frac{12}{13})^2} = \frac{1}{13}\sqrt{3600 + 144} = \frac{\sqrt{3744}}{13} = \frac{12\sqrt{26}}{13}$ units.