Given $p, q, r$ are prime numbers, and $n = pqr = 398585546147$, how do I obtain $\phi(pqr) \bmod 32$?
I tried equating it to $(p - 1)(q - 1)(r - 1) \bmod 32$ but I do not know how to proceed from there. Calculators are not allowed for this question as well.
Partial answer . . .
Although it was not explicitly specified, I suspect it was intended that the primes $p,q,r$ are distinct. Thus, in what follows, I'll assume $p,q,r$ are distinct.
As you noted, $n = pqr \implies \phi(n) = (p-1)(q-1)(r-1)$.
\begin{align*} n \text{ odd}&\implies p,q,r \text{ are odd }\\[6pt] &\implies p-1,q-1,r-1 \text{ are even}\\[6pt] &\implies 8\,|\,(p-1)(q-1)(r-1)\\[6pt] &\implies \phi(n) \equiv 0 \pmod 8\\[6pt] \end{align*}
Since $n \equiv 3 \pmod 4$, either
For case $(1)$, two of $p-1,q-1,r-1$ are multiples of $4$, and the remaining one is even, hence $\phi(n)=0 \pmod{32}$.
For case $(2)$, each of $p-1,q-1,r-1$ is even, but not a multiple of $4$. It follows that $\phi(n)$ is a multiple of $8$, but not a multiple of $16$, hence $\phi(n) = 8 \text{ or }24 \pmod{32}$.
Thus, $\phi(n)\text{ mod }32$ is one of $0,8,24$.