**Solution:**

Given equation: $ax^3 - kx^2 + lx - m = 0$

Let $\alpha, \beta, \gamma$ and $\Delta$ be the roots of the equation.

**Sum of the roots:**

The sum of the roots of a cubic equation is given by the formula:

$\alpha + \beta + \gamma + \Delta = -\frac{b}{a}$

In this case, the coefficient of $x^2$ is $-k$, so the sum of the roots is:

$\alpha + \beta + \gamma + \Delta = \frac{k}{a}$

**Product of the roots:**

The product of the roots of a cubic equation is given by the formula:

$\alpha \beta \gamma \Delta = \frac{m}{a}$

**Minimum value of $\alpha^2 + \beta^2 + 5 + 12$:**

Let's find the minimum value of the expression $\alpha^2 + \beta^2 + 5 + 12$.

First, let's find the value of $\alpha^2 + \beta^2$.

We know that $\alpha + \beta + \gamma + \Delta = \frac{k}{a}$, so $(\alpha + \beta + \gamma + \Delta)^2 = \left(\frac{k}{a}\right)^2$.

Expanding the square, we get:

$\alpha^2 + \beta^2 + \gamma^2 + \Delta^2 + 2(\alpha \beta + \alpha \gamma + \alpha \Delta + \beta \gamma + \beta \Delta + \gamma \Delta) = \frac{k^2}{a^2}$

We also know that $\alpha \beta \gamma \Delta = \frac{m}{a}$, so we can rewrite the above equation as:

$\alpha^2 + \beta^2 + \gamma^2 + \Delta^2 + 2(\alpha \beta + \alpha \gamma + \alpha \Delta + \beta \gamma + \beta \Delta + \gamma \Delta) = \frac{k^2}{a^2} - 2\frac{m}{a}$

Now, let's rewrite the expression $\alpha^2 + \beta^2 + 5 + 12$ in terms of the above equation:

$\alpha^2 + \beta^2 + 5 + 12 = (\alpha^2 + \beta^2) + 17$

Substituting the value of $\alpha^2 + \beta^2$ from the above equation, we get:

$\alpha^2 + \beta^2 + 5 + 12 = \frac{k^2}{a^2} - 2\frac{m}{a} + 17$

Simplifying the expression, we get:

$\alpha^2 + \beta^2 + 5 + 12 = \frac{k^2}{a^2} - 2\frac{m}{a} + \frac{17a^2}{a^2}$

$\alpha^2 + \beta^2 + 5 + 12 = \frac{k^2 + 17a^2 - 2am}{a^2}$

Since we want to find the minimum value of the expression, we need to find