The graph of $y = \mathrm{f}(x)$, where $\mathrm{f}(x) = ax^2 + bx + c$, passes through the point $(10, -28)$.
When the graph of $y = \mathrm{f}(x)$ is scaled parallel to the $x$-axis by factor 2, the resulting graph passes through the point $(-5, -3)$.
It is also observed that the graph of $y = \mathrm{f}(|x|)$ passes through the point $(-5, -3)$. Find the values of the constants $a$, $b$ and $c$.

1. Form the first equation using the given point on the original graph.

Substitute $x = 10, y = -28$ into $y = \mathrm{f}(x)$:
$a(10)^2 + b(10) + c = -28$
$100a + 10b + c = -28 \quad \text{--- (1)}$

2. Form the second equation using the properties of horizontal scaling.

Scaling a graph parallel to the $x$-axis by a factor of 2 transforms the function to $y = \mathrm{f}\left(\frac{x}{2}\right)$.
The new graph passes through $(-5, -3)$. Substituting these coordinates:
$\mathrm{f}\left(\frac{-5}{2}\right) = -3$
$a\left(-\frac{5}{2}\right)^2 + b\left(-\frac{5}{2}\right) + c = -3$
$\frac{25}{4}a - \frac{5}{2}b + c = -3$
$25a - 10b + 4c = -12 \quad \text{--- (2)}$

3. Form the third equation using the absolute value transformation.

The graph of $y = \mathrm{f}(|x|)$ passes through $(-5, -3)$. Substitute these coordinates:
$\mathrm{f}(|-5|) = -3$
$\mathrm{f}(5) = -3$
$a(5)^2 + b(5) + c = -3$
$25a + 5b + c = -3 \quad \text{--- (3)}$

4. Solve the system of linear equations for $a$, $b$, and $c$.

Using a graphing calculator or manual substitution to solve the simultaneous equations (1), (2), and (3).
From (3), $c = -3 - 25a - 5b$.
Substitute $c$ into (1):
$100a + 10b + (-3 - 25a - 5b) = -28$
$75a + 5b = -25 \implies 15a + b = -5 \implies b = -15a - 5$
Substitute $c$ into (2):
$25a - 10b + 4(-3 - 25a - 5b) = -12$
$-75a - 30b = 0 \implies 5a + 2b = 0$
Substitute $b = -15a - 5$ into this simplified equation:
$5a + 2(-15a - 5) = 0$
$-25a - 10 = 0 \implies a = -\frac{2}{5}$
Substituting $a$ back to find $b$ and $c$:
$b = -15\left(-\frac{2}{5}\right) - 5 = 1$
$c = -3 - 25\left(-\frac{2}{5}\right) - 5(1) = 2$