, 2024

The three different integers selected from $1$ to $6$ and whose sum is $7$ must be the integers $1$, $2$, $4$. Thus, the vertical column contains the
integers $1$, $2$, $4$
in some order.

(Can you see why no other combination of three of the given integers has
a sum of $7$?)

The three different integers selected from $1$ to $6$ and whose sum is $11$ must be $1$, $4$, $6$
or $2$, $4$, $5$
or $2$, $3$, $6$.

If the integers in the horizontal row are $1$, $4$, $6$, then there are two integers in common
with those in the vertical column, namely $1$ and $4$.

Since there have to be five different integers used in the squares, then
there cannot be two integers in common between the two lists, and so
$1$, $4$, $6$
cannot appear in the horizontal row.

Similarly, $2$, $4$, $5$
cannot appear in the horizontal row.

Thus, the horizontal row must contain the integers $2$, $3$, $6$
with $2$ appearing in the centre
square since it is the integer in common between the two lists.

The integer not appearing in any square is $5$.

The figure shows a possible arrangement of the integers.