We look for $n = p \cdot q \cdot r$, with primes $p \le q \le r$ satisfying
$$(p+1)(q+1)(r+1) = pqr + 963. \quad (1)$$
If $p=2$, the right-hand side of (1) is odd, hence the factors $q+1, r+1$ on the left must be odd too. This implies that $p=q=r=2$, which contradicts to (1). Thus we have proved that $p \ge 3$.
Now we will show that $p=3$. Suppose on the contrary that $3 < p \le q \le r$. Then the right-hand side of (1) is not divisible by $3$. The same must be true for the product $(p+1)(q+1)(r+1)$. Consequently, all the primes $p, q, r$ are congruent to $1$ modulo $3$, and hence $(p+1)(q+1)(r+1) - pqr$ is congruent to $2 \cdot 2 \cdot 2 - 1 \cdot 1 \cdot 1 = 7$, which contradicts to $(p+1)(q+1)(r+1) - pqr = 963$. Therefore, the equality $p=3$ is established.
Putting $p=3$ into (1) we get $4(q+1)(r+1) = 3qr + 963$, which can be rewritten as $(q+4)(r+4) = 975$. In view of the prime factorization $975 = 3 \cdot 5^2 \cdot 13$ and inequalities $7 \le q+4 \le r+4$, we conclude that $q+4 \le \sqrt{975} < 32$ and hence $q+4 \in \{13, 15, 25\}$. Since $q$ is a prime, it holds that $q=11$. Then $r+4=65$, and hence $r=61$ (which is a prime indeed). Consequently, the problem has a unique solution
$$n = 3 \cdot 11 \cdot 61 = 2013.$$