Determine the value of 'a' that makes the function continuous:
$f(x)=\left\{\begin{array}{l}a x^{2}-3, x< 3 \\ a x+9, x \geq 3\end{array}\right.$

Step 1 ：We are given a piecewise function $f(x)=\left\{\begin{array}{l}a x^{2}-3, x<3 \ a x+9, x \geq 3\end{array}\right.$ and we need to find the value of 'a' that makes the function continuous.

Step 2 ：For a function to be continuous at a point, the limit of the function as x approaches that point from the left (negative side) must be equal to the limit of the function as x approaches that point from the right (positive side).

Step 3 ：In this case, we need to find the value of 'a' that makes the function continuous at x=3. This means we need to set the two parts of the function equal to each other at x=3 and solve for 'a'.

Step 4 ：Setting the two parts of the function equal to each other at x=3 gives us the equation \(9a - 3 = 3a + 9\).

Step 5 ：Solving this equation gives us the value of 'a' as 2.

Step 6 ：Final Answer: The value of 'a' that makes the function continuous is \(\boxed{2}\).