A curve $D$ has equation $y = ax + b \ln x + \frac{c}{x}, \quad x > 0,$ where $a$, $b$ and $c$ are constants. It is given that $D$ has a stationary point at $x = \frac{3}{2}$ and the tangent to $D$ at the point where $x = 1$ is $y = x - 5$. Find the values of $a$, $b$ and $c$.

1. Find the coordinates of the point on the curve at $x = 1$ using the given tangent and substitute them into the curve's equation.

At $x = 1$, the $y$-coordinate on the curve is the same as that on the tangent $y = x - 5$:
$y = 1 - 5 = -4$
Substitute $(1, -4)$ into $y = ax + b \ln x + \frac{c}{x}$:
$-4 = a(1) + b \ln(1) + \frac{c}{1}$
$a + c = -4 \quad \text{--- (1)}$

2. Differentiate the equation of the curve with respect to $x$.

$\frac{dy}{dx} = a + \frac{b}{x} - \frac{c}{x^2}$

3. Form a second equation using the gradient of the tangent at $x = 1$.

The gradient of the tangent $y = x - 5$ is $1$. Therefore, at $x = 1$, $\frac{dy}{dx} = 1$.
$a + \frac{b}{1} - \frac{c}{1^2} = 1$
$a + b - c = 1 \quad \text{--- (2)}$

4. Form a third equation using the stationary point given at $x = \frac{3}{2}$.

At a stationary point, $\frac{dy}{dx} = 0$. Substitute $x = \frac{3}{2}$ into the derivative:
$a + \frac{b}{\frac{3}{2}} - \frac{c}{\left(\frac{3}{2}\right)^2} = 0$
$a + \frac{2}{3}b - \frac{4}{9}c = 0$
Multiply by 9 to clear denominators:
$9a + 6b - 4c = 0 \quad \text{--- (3)}$

5. Solve the system of three linear equations to find $a$, $b$, and $c$.

From (1), $c = -4 - a$.
Substitute $c$ into (2):
$a + b - (-4 - a) = 1$
$2a + b + 4 = 1 \implies b = -2a - 3$
Substitute expressions for $b$ and $c$ into (3):
$9a + 6(-2a - 3) - 4(-4 - a) = 0$
$9a - 12a - 18 + 16 + 4a = 0$
$a - 2 = 0 \implies a = 2$
Using $a = 2$ to find $b$ and $c$:
$b = -2(2) - 3 = -7$
$c = -4 - 2 = -6$