We have:
$$\left\{ \begin{array}{l} pqr = n \\ (p+1)(q+1)r = n+138 \end{array} \right\} \Leftrightarrow \left\{ \begin{array}{l} pqr = n \\ pqr + (p+q)r + r = n+138 \end{array} \right\} \Leftrightarrow \left\{ \begin{array}{l} pqr = n \\ (p+q+1)r = 138 \end{array} \right\}.$$
From equation
$$(p+q+1)r = 138 = 2 \cdot 3 \cdot 23 \quad (1)$$
we will determine the possible values of $p$, $q$, $r$ and in the sequel from the equation $pqr = n$. We will find all the possible values of $n$.
Since $p$, $q$, $r$ are primes, the possible values of $r$ are $2$ or $3$ or $23$, and therefore we have the cases:
* If $r = 2$, then $p+q+1 = 69 \Leftrightarrow p+q = 68$, and hence, $(p, q$ primes$)$ we find the pairs
$$(p,q) = (7,61), (p,q) = (61,7), (p,q) = (31,37), (p,q) = (37,31).$$
Hence for the product $pqr = n$ we obtain the values:
$n = 7 \cdot 61 \cdot 2 = \mathbf{854}$ or $n = 31 \cdot 37 \cdot 2 = \mathbf{2294}$.
* If $r = 3$, then $p+q+1 = 46 \Leftrightarrow p+q = 45$, and hence:
$$(p,q) = (2,43), (p,q) = (43,2) \Rightarrow n = 2 \cdot 43 \cdot 3 = \mathbf{258}.$$
* If $r = 23$, then
$$p+q+1=6 \Leftrightarrow p+q=5 \Leftrightarrow (p,q) = (2,3) \text{ or } (p,q) = (3,2),$$
and hence $n = 2 \cdot 3 \cdot 23 = \mathbf{138}$.
Therefore the possible values of $n$ are: 138, 258, 854 και 2294.