Regular expression over alphabet A = {0 , 1, 2} for words that contain at least one 0 and don't contain 11
Well, we know our word contains a $0$, so let's start with that:

$$0$$

Now let's assume that that $0$ is the first $0$ in the word. Then there may be arbitrary $1$s and $2$s before it, but $11$ is not allowed, thus $1$ must be followed by either a $2$ or by our zero:

$$(2 + 12)^*(10 + 0)$$

Now at the end we can have arbitrary characters, except for $11$. Thus each $1$ must be followed by a $0$ or $2$, or be at the end of the word:

$$(2 + 12)^*(10 + 0)(0 + 2 + 10 + 12)^*(1 + \lambda)$$

Here $\lambda$ is the empty string. If that isn't acceptable, we can distribute over the concatenation:

$$(2 + 12)^*(10 + 0)((0 + 2 + 10 + 12)^*1 + (0 + 2 + 10 + 12)^*)$$

Or alternatively we turn our logic around:

$$(2 + 12)^*(10 + 0)(1 + \lambda)(0 + 2 + 01 + 21)^*$$

And then distribute over the earlier concatenation:

$$(2 + 12)^*(101 + 01 + 10 + 0)(0 + 2 + 01 + 21)^*$$

These are all unambiguous, by the trick of recognizing a first zero, rather than the existence of an arbitrary one.