This is a selection problem

Let's select the two values of three which are going to be the same. This can be done in 3C2 ways
=> \(\frac{3!}{(2!*1!)}\) = 3 ways

Now the digit which is going to be the same can be selected in 9 ways
(we have digits from 1 to 9 to select as digits cannot be 0)

The digit which is different can be select in 8 ways. (all numbers from 1 to 9 except the one selected for the common digits)

So, in total we have 3 * 9 * 8 = 216 ways

So, answer will be D
Hope it helps!