1. We need the maximum possible value of

for a $3\times 3$ matrix whose entries are all from $\{-1,0,1\}$.

Let the rows of $P$ be vectors $\mathbf{r}_1,\mathbf{r}_2,\mathbf{r}_3$ in $\mathbb{R}^3$. Then

is the volume of the parallelepiped formed by these row vectors.

2. By Hadamard's inequality,

Since each row has entries from $\{-1,0,1\}$, the largest possible squared length of any row is

So each row has norm at most $\sqrt{3}$. Therefore,

Since determinant is an integer, this gives

So the maximum possible determinant can only be $5$ or less.

3. Now we check whether $5$ is possible.

For equality in Hadamard's inequality, rows should be mutually orthogonal and of maximum length. But in $\{-1,0,1\}^3$, there do not exist three mutually orthogonal vectors each of length $\sqrt{3}$, because vectors with all entries $\pm1$ have dot product equal to an odd integer, so cannot be $0$. Thus determinant $5$ is impossible.

So the maximum possible determinant is at most $4$.

4. Now we construct a matrix with determinant $4$:

Compute its determinant:

Now,

and

Hence,

This gives only $3$, so let us try another matrix.

Take

Then

Compute each minor:

So,

Thus determinant $4$ is attainable.

5. Since determinant cannot exceed $4$, and $4$ is achievable, the maximum possible value is