Is the key deducible with AES CTR when we know about the plaintext, IV and the ciphertext? Simply as stated, with AES running in CTR mode, can somebody work out the key (without brute-forcing) given we know about the plaintext, IV and also the ciphertext.

Assuming we also know about the cipher block size &amp; key length.

In other words, we have the pseudo random AES CTR stream + IV, can we work out what the key is?
No, that's not possible as long as the security of AES as a block cipher is not weakened.

In CTR mode a counter is encrypted using AES, and the result of that is XOR'ed with the plaintext. The counter actually consists of a nonce part and a counter part. Both the nonce and counter are known by a possible attacker.

If anybody would just be encrypting zeros the you would get exactly the key stream. This is still secure though, because the key used for AES is protected by the block cipher itself. And, as indicated by the comments of Dave Thompson, AES itself is secure against known plaintext attacks.

Basically the same reasoning can be performed for other (well known) block cipher modes of operation; the key is protected by the block cipher, how the result is used is inconsequential to security of the key.