Show that BT is an altitude of triangle ABC and determine the coordinates of C, given A(p; 8), T(1; q), B(-2; -2), and the equations of lines AC and BC.

Point $A(p; 8)$ lies on the line $AC$ with equation $4y + 3x = 11$. Substitute $x=p$ and $y=8$ into the equation:

$$4(8) + 3p = 11$$ $$32 + 3p = 11$$ $$3p = 11 - 32$$ $$3p = -21$$ $$p = -7$$

Point $T(1; q)$ lies on the line $AC$ with equation $4y + 3x = 11$. Substitute $x=1$ and $y=q$ into the equation:

$$4q + 3(1) = 11$$ $$4q + 3 = 11$$ $$4q = 11 - 3$$ $$4q = 8$$ $$q = 2$$

Thus, $p = -7$ and $q = 2$.

An altitude is perpendicular to the opposite side. We need to show that $BT \perp AC$.
First, find the gradient of $BT$ using points $B(-2; -2)$ and $T(1; 2)$:

$$m_{BT} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - (-2)}{1 - (-2)} = \frac{4}{3}$$

Next, find the gradient of $AC$ from its equation $4y + 3x = 11$. Rearrange to $y = mx + c$ form:

$$4y = -3x + 11$$ $$y = -\frac{3}{4}x + \frac{11}{4}$$

So, the gradient of $AC$ is $m_{AC} = -\frac{3}{4}$.
Now, check if $BT$ and $AC$ are perpendicular by multiplying their gradients:

$$m_{BT} \times m_{AC} = \frac{4}{3} \times \left(-\frac{3}{4}\right) = -1$$

Since the product of their gradients is $-1$, $BT \perp AC$. Therefore, BT is an altitude of $\triangle ABC$.

Point $C$ is the intersection of line $AC$ and line $BC$.
Equation of $AC$: $4y + 3x = 11$ (1)
Equation of $BC$: $7y - x = -12$ (2)
From equation (2), express $x$ in terms of $y$:

$$x = 7y + 12$$

Substitute this expression for $x$ into equation (1):

$$4y + 3(7y + 12) = 11$$ $$4y + 21y + 36 = 11$$ $$25y + 36 = 11$$ $$25y = 11 - 36$$ $$25y = -25$$ $$y = -1$$

Substitute $y = -1$ back into the expression for $x$:

$$x = 7(-1) + 12$$ $$x = -7 + 12$$ $$x = 5$$

The coordinates of C are $\boxed{(5; -1)}$.

We can use the formula for the area of a triangle: Area $= \frac{1}{2} \times \text{base} \times \text{height}$. Since $BT$ is an altitude to $AC$, we can use $AC$ as the base and $BT$ as the height.
First, calculate the length of $AC$ using points $A(-7; 8)$ and $C(5; -1)$:

$$AC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ $$AC = \sqrt{(5 - (-7))^2 + (-1 - 8)^2}$$ $$AC = \sqrt{(12)^2 + (-9)^2}$$ $$AC = \sqrt{144 + 81}$$ $$AC = \sqrt{225}$$ $$AC = 15 \text{ units}$$

Next, calculate the length of $BT$ using points $B(-2; -2)$ and $T(1; 2)$:

$$BT = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ $$BT = \sqrt{(1 - (-2))^2 + (2 - (-2))^2}$$ $$BT = \sqrt{(3)^2 + (4)^2}$$ $$BT = \sqrt{9 + 16}$$ $$BT = \sqrt{25}$$ $$BT = 5 \text{ units}$$

Now, calculate the area of $\triangle ABC$:

$$\text{Area} = \frac{1}{2} \times AC \times BT$$ $$\text{Area} = \frac{1}{2} \times 15 \times 5$$ $$\text{Area} = \frac{1}{2} \times 75$$ $$\text{Area} = 37.5 \text{ square units}$$

The area of $\triangle ABC$ is $\boxed{37.5 \text{ square units}}$.