We are given a $7 \times 7$ array. We want to remove a $1 \times 1$ square, such that the remaining shape can be covered with $1 \times 3$ triomino.
What are the possible positions of the eliminated square?
We are given a $7 \times 7$ array. We want to remove a $1 \times 1$ square, such that the remaining shape can be covered with $1 \times 3$ triomino.
What are the possible positions of the eliminated square?
You can color the array as $$ \pmatrix{ 0 & 1 & 2 & 0 & 1 & 2 & 0 \\ 1 & 2 & 0 & 1 & 2 & 0 & 1 \\ 2 & 0 & 1 & 2 & 0 & 1 & 2 \\ 0 & 1 & 2 & 0 & 1 & 2 & 0 \\ 1 & 2 & 0 & 1 & 2 & 0 & 1 \\ 2 & 0 & 1 & 2 & 0 & 1 & 2 \\ 0 & 1 & 2 & 0 & 1 & 2 & 0 \\ } $$ and since there are $17$ zeros and $16$ of each other number, the missing square has to be a $0$.
But you can also color it as $$ \pmatrix{ 0 & 1 & 2 & 0 & 1 & 2 & 0 \\ 2 & 0 & 1 & 2 & 0 & 1 & 2 \\ 1 & 2 & 0 & 1 & 2 & 0 & 1 \\ 0 & 1 & 2 & 0 & 1 & 2 & 0 \\ 2 & 0 & 1 & 2 & 0 & 1 & 2 \\ 1 & 2 & 0 & 1 & 2 & 0 & 1 \\ 0 & 1 & 2 & 0 & 1 & 2 & 0 \\ } $$ so the missing quare has to be $0$ in both colorings, that is, it has to be a corner or a midpoint of a side, or the center point.
It is easy to see how to do the tiling in each of these cases:
If a corner is missing, say the upper left corner, then place to triminoes horizontally covering the top, and the rest vertically, two in each column.
If a midpoint of a side is missing, do the same, only now the two horizontal triminoes are separated by the missing space.
if the center square is missing, then place two triminoes horizontally across the center row, separated by the center space. You are left ith two 3x7 horizontal bands, which can be tiled vertically.
Tip: Work out what blocks can be placed in the grid: $1 \times 3, 4 \times 3, 7 \times 3, 5 \times 3, 6/times 3 $ etc.
Tip 2: Use Trial and error:
Tip 2.1: Eliminate parts of the grid with $7 \times x$ blocks to make the grid smaller to work with. This methods workings can be easily proved
Tip 2.1.1 : Don't crop smaller than a $4 \times 4$ grid
Tip X:
If you haven't noticed already, symmetries and rotations are equivalent to each other.
The answer I got was $9$. Since I believe that you haven't attempted this yourself, I will leave the proof to you, using the tips above...