From the question, we have 3 kinds of meat and 4 kinds of vegatables.

To get a two-distinct-meat pizza, we need to choose the first meat and the second meat.

From 3 kinds of meat, to choose the first meat, we have 3 options. After choosing the first meat, we have 2 kinds left (because the first and the second meat need to be distinct), so we have 2 options for the second meat.

So, we have  \(3\times2=6\) two-distinct-meat pizzas.

To get a two-distinct-vegatable pizza, we need to choose the first vegatable and the second vegatable.

From 4 kinds of vegatables, to choose the first vegatable, we have 4 options. After choosing the first vegatable, we have 3 kinds left (because the first and the second vegatable need to be distinct), so we have 3 options for the second vegatable.

So, we have \(4\times3=12\) two-distinct-vegatable pizzas.

To get a one-meat-one-vegatable pizza, we need to choose the meat and the vegatable.

From 3 kinds of meat, we have 3 options. From 4 kinds of vegatables, we have 4 options.

So, we have  \(3\times4=12\) one-meat-one-vegatable pizzas.

After all, we have \(6+12+12=30\) combinations of pizza.