In How to get $P(A\mid B, C)$ from $P(A\mid B)$, $P(A\mid C)$, $P(B,C)$ it is shown that there isn't enough information to compute $P(A\mid BC)$ from $P(A\mid B)$, $P(A\mid C)$ and $P(B\mid C)$.
My question: what other information should we provide?
Motivation
Consider the associated Bernoulli distributions $B_A$, $B_B$ and $B_C$ (meaning the B.D. with parameters $P(A)$, etc.)
It is easy to see that the (Pearson) correlation coefficient $\rho_{B_A,B_B}$ conveys the same information that $P(A|B)$ (you can compute one from the other). This fact can be used in software to produce a sample of $B$ conditional to a sample of $A$ by generating two samples $a_1,\cdots,a_n$ and $b_1,\cdots,b_n$ appropriately correlated.
So, my question could be rephrased as: Is it possible to compute $\rho_{B_A,B_{BC}}$ from $\rho_{B_A,B_B}$, $\rho_{B_A,B_C}$ and $\rho_{B_B,B_C}$ assuming you also know $B_A$, $B_B$ and $B_C$?
Actually I don't see this, unless you are implicitly assuming that $P(A)$ and $P(B)$ are known when you write that $B_A$ and $B_B$ are Bernoulli distributions with parameters $P(A)$ and $P(B).$
The question How to get $P(A\mid B, C)$ from $P(A\mid B)$, $P(A\mid C)$, $P(B,C)$ did not assume $P(A),$ $P(B),$ and $P(C)$ were known, and neither did the answer. I think you are looking for an answer that assumes those probabilities are known.
So let's suppose you know not just $P(A\mid B),$ $P(A\mid C),$ and $P(B\mid C),$ but also $P(A),$ $P(B),$ and $P(C),$ and that each of the probabilities $P(A),$ $P(B),$ and $P(C)$ is non-zero. Equivalently, we can say you know $P(A),$ $P(B),$ $P(C),$ $P(A\cap B),$ $P(A\cap C),$ and $P(B\cap C),$ since you can compute the first six probabilities from these last six probabilities and vice versa.
Now let's define a few symbols for convenience: \begin{align} p &= P(A \cap B^\complement \cap C^\complement), \\ q &= P(A^\complement \cap B \cap C^\complement), \\ r &= P(A^\complement \cap B^\complement \cap C), \\ s &= P(A \cap B \cap C^\complement), \\ t &= P(A \cap B^\complement \cap C), \\ u &= P(A^\complement \cap B \cap C), \\ v &= P(A \cap B \cap C). \end{align}
Then what we know is \begin{align} p + s + t + v &= P(A), \\ q + s + u + v &= P(B), \\ r + t + u + v &= P(C), \\ s + v &= P(A\cap B), \\ t + v &= P(A\cap C), \\ u + v &= P(B\cap C). \end{align}
We have only six equations for seven unknowns. We also have some constraints: $p \geq 0,$ $q \geq 0,$ $r \geq 0,$ $s \geq 0,$ $t \geq 0,$ $u \geq 0,$ $v \geq 0,$ and $p + q + r + s + t + u + v \leq 1,$ but that's still not enough to determine a unique solution in general.
What this means is that in the general case, you don't know $P(A \cap B \cap C)$; at best you can find a minimum and maximum possible value. But if it were possible to deduce any of the various three-variable conditional probabilities such as $P(A\mid B \cap C),$ then from the known probabilities you could also compute $P(A \cap B \cap C)$ exactly.
As an example, suppose the probability space contains eight equally probable, disjoint events $a, b, c, d, e, f, g,$ and $h,$ and consider the events \begin{align} A &= \{a, d, e\}, \\ B &= \{b, d, f\}, \\ C &= \{c, e, f\}, \\ A' &= \{a, d, g\}, \\ B' &= \{b, e, g\}, \\ C' &= \{c, f, g\}. \end{align} Then $P(A) = P(A') = P(B) = P(B') = P(C) = P(C') = \frac38$ and $P(A\cap B) = P(A'\cap B') = P(A\cap C) = P(A'\cap C') = P(B\cap C) = P(B'\cap C') = \frac18,$ but $P(A\mid B \cap C) = 0$ while $P(A'\mid B' \cap C') = 1.$
In conclusion, even if you know all of the probabilities $P(A),$ $P(B),$ $P(C),$ $P(A\cap B),$ $P(A\cap C),$ $P(B\cap C),$ $P(A\mid B),$ $P(B\mid A),$ $P(A\mid C),$ $P(C\mid A),$ $P(C\mid B),$ and $P(B\mid C),$ in the general case you still do not have enough information to determine other probabilities such as $P(A\mid B \cap C)$ or $P(A\cap B \mid C).$