Given $4$ variables and $5$ pairwise products, find the $6$th pairwise product?

You can separate the six products into three pairs with each pair having different factors $$x_1\cdot x_2\quad \ x_3\cdot x_4\\ x_1\cdot x_3\quad \ x_2 \cdot x_4\\ x_1 \cdot x_4\quad \ x_2 \cdot x_3$$ When we multiply the partial products on each line, we should get the same result. The only two pairs that have the same product are $2 \cdot 6$ and $3 \cdot 4$, so the product of the last line must also be $12$. The sixth partial product is $$\frac {12}5$$ Now we can check that the solution works. By symmetry we can assign the first line $2 \cdot 6,$ the second $3 \cdot 4$ and the last $5 \cdot \frac {12}5$ but we cannot be sure of the order of the last.Then $\frac {x_3}{x_2}=\frac 32.$ If the last is $5 \cdot \frac {12}5$ then $\frac {x_4}{x_2}=\frac 52, \frac {x_1}{x_2}=\frac 54$. The product of them all is $12$, so we have $$\frac 32\cdot \frac 52 \cdot \frac 54 x_2^4=12\\x_2=\sqrt{\frac 85}\\ x_1=\frac 54\sqrt {\frac 85}\\x_3=\frac 32\sqrt{\frac 85}\\x_4=\frac 52 \sqrt{\frac 85}$$ If we switch the products in the last line, we get another solution $$x_2=\sqrt{\frac {10}3}\\ x_1=\sqrt {\frac 65}\\x_3=\sqrt{\frac{15}2}\\x_4=\sqrt{\frac{24}5}$$ We can permute the assignment of the variables at will.

We have the linear system $$ \left( {\matrix{ 1 & 1 & 0 & 0 \cr 1 & 0 & 1 & 0 \cr 1 & 0 & 0 & 1 \cr 0 & 1 & 1 & 0 \cr 0 & 1 & 0 & 1 \cr 0 & 0 & 1 & 1 \cr } } \right)\left( {\matrix{ {\ln x_{\,1} } \cr {\ln x_{\,2} } \cr {\ln x_{\,3} } \cr {\ln x_{\,4} } \cr } } \right) = \left( {\matrix{ {\ln p_{\,1} } \cr {\ln p_{\,2} } \cr {\ln p_{\,3} } \cr {\ln p_{\,4} } \cr {\ln p_{\,5} } \cr {\ln p_{\,6} } \cr } } \right)\quad \Rightarrow \quad {\bf M}\;{\bf y} = {\bf p} $$ The rank of the coefficient matrix is $4$, and so the rank of the augmented matrix shall be not greater than $4$.
This pose a constraint on the values and on the order of the $p_k$.

We can have a better look to the situation if we perform a Gaussian elimination on the augmented matrix.
We do that by performing a LU decomposition of $\bf M$ and then right-multiplying by the inverse of the L component, and arrive to $$ \left( {\matrix{ 1 & 1 & 0 & 0 \cr 0 & { - 1} & 1 & 0 \cr 0 & 0 & { - 1} & 1 \cr 0 & 0 & 0 & 2 \cr 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 \cr } } \right)\left( {\matrix{ {\ln x_{\,1} } \cr {\ln x_{\,2} } \cr {\ln x_{\,3} } \cr {\ln x_{\,4} } \cr } } \right) = \left( {\matrix{ 1 & 0 & 0 & 0 & 0 & 0 \cr { - 1} & 1 & 0 & 0 & 0 & 0 \cr 0 & { - 1} & 1 & 0 & 0 & 0 \cr { - 1} & { - 1} & 2 & 1 & 0 & 0 \cr 0 & 1 & { - 1} & { - 1} & 1 & 0 \cr 1 & 0 & { - 1} & { - 1} & 0 & 1 \cr } } \right)\left( {\matrix{ {\ln p_{\,1} } \cr {\ln p_{\,2} } \cr {\ln p_{\,3} } \cr {\ln p_{\,4} } \cr {\ln p_{\,5} } \cr {\ln p_{\,6} } \cr } } \right) $$

It is evident that the system is solvable, and has a unique solution, if and only if the $p_k$'s satisfy the last two equations, i.e. if $$ \left\{ \matrix{ \ln p_{\,2} + \ln p_{\,5} = \ln p_{\,3} + \ln p_{\,4} \hfill \cr \ln p_{\,1} + \ln p_{\,6} = \ln p_{\,3} + \ln p_{\,4} \hfill \cr} \right.\quad \Rightarrow \quad p_{\,1} p_{\,6} = p_{\,2} p_{\,5} = p_{\,3} p_{\,4} $$ and this gives a general method to solve this type of problems.

With the data you give, the only way to have two couples with the same product is $2 \cdot 6 = 3 \cdot 4$, and therefore $$ 2 \cdot 6 = {{12} \over 5} \cdot 5 = 3 \cdot 4 $$ which means $$ {\bf p}^{\,T} = \left( {\ln 2,\ln \left( {{{12} \over 5}} \right),\ln 3,\ln 4,\ln 5,\ln 6} \right) $$ Then the upper four equations are readily solved to give $$ {\bf y} = {1 \over 2}\left( {\matrix{ {\ln {6 \over 5}} \cr {\ln {{10} \over 3}} \cr {\ln {{24} \over 5}} \cr {\ln {{15} \over 2}} \cr } } \right)\quad \Rightarrow \quad {\bf x} = \left( {\matrix{ {\sqrt {{6 \over 5}} } \cr {\sqrt {{{10} \over 3}} } \cr {\sqrt {{{24} \over 5}} } \cr {\sqrt {{{15} \over 2}} } \cr } } \right) $$ and in fact $$ \eqalign{ & \sqrt {{6 \over 5}} \sqrt {{{10} \over 3}} = 2\quad \sqrt {{6 \over 5}} \sqrt {{{24} \over 5}} = {{12} \over 5}\quad \sqrt {{6 \over 5}} \sqrt {{{15} \over 2}} = 3 \cr & \sqrt {{{10} \over 3}} \sqrt {{{24} \over 5}} = 4\quad \sqrt {{{10} \over 3}} \sqrt {{{15} \over 2}} = 5\quad \sqrt {{{24} \over 5}} \sqrt {{{15} \over 2}} = 6 \cr} $$

However, swapping the values of $p_1$ and $p_6$ we get a different quadruple as result $$ {\bf x} = \left( {\matrix{ {\sqrt {{{18} \over 5}} } \cr {\sqrt {10} } \cr {\sqrt {{8 \over 5}} } \cr {\sqrt {{5 \over 2}} } \cr } } \right) $$ which checks as well.

I did not check for the other allowed permutations of $\bf p$.

Suppose the four numbers are $0<a<b<c<d$ Note: we can assume they are distinct since no products are repeated.

Let the six products be $P_1≤P_2≤\cdots ≤P_6$.

It is easy to see that these must be of the form $$ab<ac<\{bc,ad\}<bd<cd$$

Where the order of the two terms in the middle is uncertain.

If we assume that $P_6>6$ then we get $$\frac {P_1\times P_6}{P_2}=P_5\implies \frac {2P_6}{3}=6\implies P_6=9$$

But there are no solutions for $(a,b,c,d)$ consistent with this (brute force).

Similarly we can not have $P_1<2$.

If we had $P_2$ as the missing term then we'd get $$\frac {2\times 6}{5}=P_2\implies \boxed {P_2=\frac {12}5}$$

That one works! Indeed we could have $$a=2\sqrt {\frac 25}\quad b=\sqrt {\frac 52}\quad c= 3\sqrt {\frac25}\quad d=\sqrt {10}$$

I did not try to analyze the other cases, though this would be no harder.