This is related to Weibel, Introduction to Homological Algebra, Definition 2.7.7 on pg 64.
An left exact functor $T$ with $p$ variables is called right balanced functor if the following holds.
(1) When any one of covariant variables of $T$ is replaced by an injective, then $T$ becomes an exact functor of remaining variables.
(2) When any one of contravariant variables of $T$ is replaced by a projective, then $T$ becomes an exact functor of remaining variables.
The book gave example as $Hom(A\otimes B, C)$ over some ring $R$. I am aware that $Hom(A,C)$ is right balanced but I could not see $Hom(A\otimes B,C)$ is right balanced.
$\textbf{Q:}$ Why is $Hom(A\otimes B,C)$ right balanced? Consider $A$ projective and choose $B$ s.t. $A\otimes B$ non-projective. Then $Hom(A\otimes B,-)$ can't be an exact functor. Take $R=Z, A=Z$ and $B=Z_2$. Then one has to check $Hom(Z_2,-)$ is not exact for some short exact sequence. Consider $0\to Z\xrightarrow{\times 2}Z\to Z_2\to 0$. Apply $Hom(Z_2,-)$ functor. By $Hom(Z_2,Z)=0$, I have $0\to 0\to 0\to Z_2$ where the last map can't be surjection.
My guess the book should mean $Ext^i(Tor^j(A,B),C)$ instead.
I think Weibel is wrong that Hom$(A\otimes B,C)$ is right balanced. Cartan-Eilenberg give the same definition, but they say they know of no example of a balanced functor in three variables.
(Edited to completely change my answer a few minutes after I posted it.)