Let the magic square be $\begin{bmatrix}a & b & c & d \\ x_1 & x_2 & x_3 & x_4 \\ x_5 & x_6 & x_7 & x_8 \\ x_9 & x_{10} & x_{11} & x_{12}\end{bmatrix}$ where $a,b,c,d$ represents the date.
The conditions that the rows, columns, and diagonals sum to $a+b+c+d$ gives us $9$ linear equations for $12$ variables. The resulting system has rank $8$.
Therefore, any solution can be obtained by taking a particular solution and adding a linear combination of $12-8 = 4$ "basis vectors" for the nullspace of the system.
The matrix $\begin{bmatrix}a & b & c & d \\ c & d & a & b \\ d & c & b & a \\ b & a & d & c\end{bmatrix}$ is a magic square with the correct top row.
The following $4$ matrices form a basis of the nullspace of the system (all rows, cols, diags sum to $0$):
$\begin{bmatrix}0 & 0 & 0 & 0\\ 2 & -1 & -1 & 0 \\ -2 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0\end{bmatrix}$, $\begin{bmatrix}0 & 0 & 0 & 0\\ 1 & 0 & 0 & -1 \\ -1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0\end{bmatrix}$, $\begin{bmatrix}0 & 0 & 0 & 0\\ 1 & 0 & -1 & 0 \\ -1 & 1 & 0 & 0 \\ 0 & -1 & 1 & 0\end{bmatrix}$, $\begin{bmatrix}0 & 0 & 0 & 0\\ 2 & -1 & 0 & -1 \\ -1 & 1 & 0 & 0 \\ -1 & 0 & 0 & 1\end{bmatrix}$.
Take the first matrix and add any linear combination of the above four matrices to get a solution.
Why it's impossible:
There are actually 4 more arithmetic sequences that will always be present in a magic square.
Let the square:
$$a,b,c$$
$$d,e,f$$
$$g,h,i$$
So from $(3)$ and $e=\frac{r}{3}$ we get the equations:
$$a-e=e-i$$
$$b-e=e-h$$
$$c-e=e-g$$
$$d-e=e-f$$
$(6)$ From these we get:
$$2e=a+i=b+h=c+g=d+f$$
Next take:
$$r=a+b+c=a+d+g\implies b+c=d+g$$
Then from $(6)$, substitute in $g=b+h-c$ to get:
$$b+c=d+(b+h-c)$$
Rearrange:
$$d-c=c-h$$
The same "trick" can be done too for $(b,g,f),(b,i,d),(h,a,f)$ either using the same method or the symmetries of the magic square.
$(7)$ We have:
$$d-i=i-b$$
$$b-g=g-f$$
$$f-a=a-h$$
$$h-c=c-d$$
Note the middle elements are the corners, which is helpful for remembering these.
Back to your question now.
For $r=15$ we have 4 sequences across the center: $(1,9),(2,8),(3,7),(4,6)$
From these we can make 5 more arithmetic sequences: $(1,2,3),(2,4,6),(1,4,7),(3,6,9),(7,8,9)$
All 5 progressions have at least one even number so no such square is possible!
Finding the smallest odd magic square:
Since the square with $5$ in the center (and a total of $15$) is impossible. Let's try the square with $7$ in the center.
The sequences about 7 are $(1,13),(3,11),(5,9)$
From these we can pull the sequences $(1,3,5),(1,5,9),(9,11,13),(5,9,13)$
The middle elements will become the centers and the left/right elements will become the middles of the edges of the square. This implies $5$ and $9$ will be duplicated. So no distinct square exists with center element $e=7$
However for $e=9$ we can take the pairs about $9$: $(1,17),(3,15),(5,13),(7,11)$
Now note for the equations in $(7)$, each equation's last variable is the first variable of the equation following it. That is to say, we must pull 4 sequences from the elements of the pairs above such that they form a sort of loop with their first and last elements. The right sequence can be found with a small bit of work:
$$\rightarrow(1,7,13)\rightarrow(13,15,17)\rightarrow(17,11,5)\rightarrow(5,3,1)\rightarrow$$
Now we construct the square, starting with $(1,7,13)$:
$$a,[1],c$$
$$[13],[9],f$$
$$g,h,[7]$$
Then $(13,15,17)$:
$$a,1,[15]$$
$$[13],9,f$$
$$g,[17],7$$
Then $(17,11,5)$:
$$[11],1,15$$
$$13,9,[5]$$
$$g,[17],7$$
Finally $(5,3,1)$:
$$11,[1],15$$
$$13,9,[5]$$
$$[3],17,7$$
And we're done. The smallest all odd magic square has $r=27$
$$11,1,15$$
$$13,9,5$$
$$3,17,7$$
Best Answer
Let's say we put in the numbers $a$, $b$, and $c$: $$\begin{pmatrix} a & 3 & 6\\ 5 & b & 5\\ 4 & 7 & c \end{pmatrix}$$ The sums of the rows are $a+9$, $b+10$, and $c+11$. The sums of the columns are the same (indeed, if that were not the case, it'd be impossible to make it into a magic square). Thus, we want the numbers $a$, $b$, and $c$ to satisfy $$a+9=b+10=c+11.$$ Such triples of numbers are precisely those of the form $a=x$, $b=x-1$, and $c=x-2$ for some number $x$. But we also want the diagonals to add up to the same value; thus, we want $$a+b+c=4+b+6$$ $$x+(x-1)+(x-2)=4+(x-1)+6$$ $$3x-3=x+9$$ $$x=6$$ Thus, the unique entries we can put in to make the matrix a magic square are $$\begin{pmatrix} \fbox{6} & 3 & 6\\ 5 & \fbox{5} & 5\\ 4 & 7 & \fbox{4} \end{pmatrix}$$