There are 18 plaintext and ciphertext letters $p_j$ and $c_j$, $0\le j<18$ (with $j<6$ for the "first plaintext"), all of which are known except $p_7..p_{17}$.
Let $M=\pmatrix{m_{0,0}&m_{0,1}&m_{0,2}\\m_{1,0}&m_{1,1}&m_{1,2}\\m_{2,0}&m_{2,1}&m_{2,2}}$ be the key matrix (unknown, except that it is invertible).
We have 18 linear equations in $\mathbb{Z}_{26}$
$$c_j=m_{j\bmod3,0}\cdot p_{3{\lfloor{j/3}\rfloor}}+m_{j\bmod3,1}\cdot p_{3{\lfloor{j/3}\rfloor}+1}+m_{j\bmod3,2}\cdot p_{3{\lfloor{j/3}\rfloor}+2}$$
with 20 unknowns, and the tiny information that $M$ is invertible. By an entropy argument, this can't be solved in the general case unless by exploiting redundancy in the second plaintext.
One (naïve) possibility to solve the problem could be: using a computer, for each of the $26\cdot26=676$ combinations of $p_7$ and $p_8$, solve the first 9 equations (when possible and there is a unique invertible solution for $M$, which is I guess is for most combinations of $p_7$ and $p_8$), and display the resulting second ciphertext $p_6..p_{17}$; then find the most likely one using vision and brain.
Update: But wait, we should use as unknown the decryption matrix $\hat M$ and the equations
$$p_j=\hat m_{j\bmod3,0}\cdot c_{3{\lfloor{j/3}\rfloor}}+\hat m_{j\bmod3,1}\cdot c_{3{\lfloor{j/3}\rfloor}+1}+\hat m_{j\bmod3,2}\cdot c_{3{\lfloor{j/3}\rfloor}+2}$$
The 3 unknowns $\hat m_{0,0},\hat m_{0,1},\hat m_{0,2}$ can (likely) be found just by solving the system of 3 equations involving $p_0, p_3, p_6$. We can then compute $p_9, p_{12}, p_{15}$ without any guesswork.
Similarly, any guess of any of the 8 remaining plaintext letters $p_j$ gives one extra equation involving $\hat m_{j\bmod3,0},\hat m_{j\bmod3,1},\hat m_{j\bmod3,2}$, which (likely) is enough to deduce these 3 unknowns and 3 others plaintext letters. That greatly ease tabulating the possible plaintexts, from which only a few will hopefully emerge as making sense. That could even be workable just with pencil and paper.
In a computer search, the possible plaintexts could be ranked by their likelyhood given the frequency of digrams in English text.
Further update: It could well be that the system of equations for $\hat m_{0,0},\hat m_{0,1},\hat m_{0,2}$ has several solutions (I guess 2, 13, or 26), making the problem harder. It could also be that this system has no solution, in which case we could rule out the statement as faulty.