Solutions obtained by accurate drawing will not be accepted. The coordinates of points A, B and C are $(-2,3)$, $(2,-4)$ and $(7,11)$ respectively.

1. Apply the shoelace formula to find the area of the triangle

$\text{Area} = \frac{1}{2} \left| \begin{matrix} -2 & 2 & 7 & -2 \\ 3 & -4 & 11 & 3 \end{matrix} \right|$

2. Evaluate the determinant to find the final area

$\text{Area} = \frac{1}{2} | ((-2)(-4) + (2)(11) + (7)(3)) - ((3)(2) + (-4)(7) + (11)(-2)) |$
$= \frac{1}{2} | (8 + 22 + 21) - (6 - 28 - 22) |$
$= \frac{1}{2} | 51 - (-44) |$
$= \frac{1}{2} (95) = 47.5$

1. Calculate the gradient of the line segment AB

Gradient of $AB = \frac{-4 - 3}{2 - (-2)} = -\frac{7}{4}$

2. Calculate the gradient of the line segment AD

Gradient of $AD = \frac{-2 - 3}{-1 - (-2)} = \frac{-5}{1} = -5$

3. Compare the gradients to determine collinearity

Since the gradient of $AB \neq$ gradient of $AD$, the points $A$, $B$ and $D$ do not lie on the same straight line. Therefore, they are not collinear.

1. Find the gradient of the line $BC$.

Gradient of $BC = \frac{11 - (-4)}{7 - 2} = \frac{15}{5} = 3$

2. Find the equation of the line $BC$.

$y - (-4) = 3(x - 2)$
$y + 4 = 3x - 6$
$y = 3x - 10$

3. Find the gradient and equation of the perpendicular line from $A$ to $BC$.

Gradient of perpendicular line $= -\frac{1}{3}$
Equation of perpendicular line passing through $A(-2, 3)$:
$y - 3 = -\frac{1}{3}(x - (-2))$
$3y - 9 = -x - 2$
$x + 3y = 7$

4. Solve the equations simultaneously to find the coordinates of the foot of the perpendicular.

Substitute $y = 3x - 10$ into $x + 3y = 7$:
$x + 3(3x - 10) = 7$
$x + 9x - 30 = 7$
$10x = 37$
$x = 3.7$

Substitute $x = 3.7$ into $y = 3x - 10$:
$y = 3(3.7) - 10 = 11.1 - 10 = 1.1$

The coordinates of the foot of the perpendicular are $(3.7, 1.1)$.