The point $A$ has coordinates $(1.5, 0, 2)$, and the planes $\pi_1$ and $\pi_2$ have equations $x + y = 4$ and $3x + 2y - 5z = 7$ respectively.

1. Find the direction vector of the line of intersection by taking the cross product of the normal vectors of the two planes.

Let the normal vectors to $\pi_1$ and $\pi_2$ be $\mathbf{n}_1$ and $\mathbf{n}_2$ respectively.
$\mathbf{n}_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$, $\mathbf{n}_2 = \begin{pmatrix} 3 \\ 2 \\ -5 \end{pmatrix}$

The direction vector $\mathbf{d}$ of the line $l$ is orthogonal to both normal vectors, so $\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2$.

$\mathbf{d} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 3 \\ 2 \\ -5 \end{pmatrix} = \begin{pmatrix} (1)(-5) - (0)(2) \\ (0)(3) - (1)(-5) \\ (1)(2) - (1)(3) \end{pmatrix} = \begin{pmatrix} -5 \\ 5 \\ -1 \end{pmatrix}$

2. Find a point on the line of intersection by setting $z=0$ and solving the resulting linear system.

Let $z = 0$. The equations of the planes become:
$x + y = 4 \quad \text{--- (1)}$
$3x + 2y = 7 \quad \text{--- (2)}$

From (1), $y = 4 - x$. Substituting this into (2):
$3x + 2(4 - x) = 7$
$3x + 8 - 2x = 7$
$x = -1$

Substitute $x = -1$ back into $y = 4 - x$:
$y = 4 - (-1) = 5$

Therefore, a point on the line $l$ is $(-1, 5, 0)$.

3. State the vector equation of the line $l$.

Using the position vector of the point $(-1, 5, 0)$ and the direction vector $\begin{pmatrix} -5 \\ 5 \\ -1 \end{pmatrix}$:

$\mathbf{r} = \begin{pmatrix} -1 \\ 5 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} -5 \\ 5 \\ -1 \end{pmatrix}, \lambda \in \mathbb{R}$

1. Formulate the vector equation of the line passing through $A$ and perpendicular to $\pi_1$.

The normal vector to the plane $\pi_1: x + y = 4$ is $\mathbf{n} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$.

The line passing through $A(1.5, 0, 2)$ and perpendicular to $\pi_1$ is:
$l: \mathbf{r} = \begin{pmatrix} 1.5 \\ 0 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \text{ for } \lambda \in \mathbb{R}$

2. Substitute the line equation into the equation of $\pi_1$ to solve for the parameter $\lambda$.

Since the foot of the perpendicular lies on $\pi_1$, its coordinates must satisfy $x + y = 4$.
$(1.5 + \lambda) + (0 + \lambda) = 4$
$1.5 + 2\lambda = 4$
$2\lambda = 2.5 \implies \lambda = 1.25$

3. Calculate the position vector and hence coordinates of the foot of the perpendicular.

Substitute $\lambda = 1.25$ back into the line equation:
$\vec{OF} = \begin{pmatrix} 1.5 \\ 0 \\ 2 \end{pmatrix} + 1.25 \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2.75 \\ 1.25 \\ 2 \end{pmatrix}$

The coordinates of the foot of the perpendicular are $(2.75, 1.25, 2)$.