To ensure the function $f(x)$ is continuous, we need to guarantee that the function values match at the point where $x = 1$, since the function is defined differently for $x < 1$ and $x \geq 1$.

Step-by-step solution:

1. Left-hand limit $x \to 1^-$: For $x < 1$, $f(x) = 3 - x^2$. $\lim_{x \to 1^-} f(x) = 3 - (1)^2 = 3 - 1 = 2$

2. Right-hand limit $x \to 1^+$: For $x \geq 1$, $f(x) = ax^2 - 1$. $\lim_{x \to 1^+} f(x) = a(1)^2 - 1 = a - 1$

3. Continuity condition: For continuity at $x = 1$, the left-hand limit must equal the right-hand limit: $2 = a - 1$

4. Solve for $a$: $a = 2 + 1 = 3$

Thus, the value of $a$ that makes the function $f(x)$ continuous is $\boxed{3}$.

Do you want further details or have any questions? Here are five related problems to expand your understanding:

1. What happens if we try a different value of $a$? Is there any discontinuity?
2. How would you check the continuity of a piecewise function if there were three pieces?
3. What are the conditions for differentiability at $x = 1$ for the same function?
4. How can you generalize this problem to other points of discontinuity?
5. How would you extend this function to ensure it's smooth (differentiable) everywhere?

Tip: To ensure a function is continuous, always verify that the left-hand and right-hand limits are equal at the point of interest.