This is the question:
Approximately how much water should be added to 10.0 mL of 12.0 M HCl so that it has the same pH as 0.90 M acetic acid (ka=1.80 * 10^-5)?

I am unsure of how to do this

Here is my attempt at this problem. Am I right?

Once you have calculated the pH of the acetic acid then it means you know concentration of H+ = 10^-pH
New concentration is 10^-pH for HCl .
Use dilution law to get volume needed to dilute.
M1V1= M2V2