Let the given two sides of the parallelogram lie along the lines $L_1: x+y=3$ and $L_2: x-y+3=0 \iff x-y=-3.$

Since two sides of a parallelogram are along these lines, they must meet at one vertex of the parallelogram.

Find their point of intersection. Solve $x+y=3$ $x-y=-3$ Adding, $2x=0 \implies x=0.$ Then $y=3.$

So one vertex is $A=(0,3).$

In a parallelogram, the diagonals bisect each other. Hence if the diagonals intersect at $(2,4)$, then $(2,4)$ is the midpoint of each diagonal.

Therefore the vertex opposite to $A=(0,3)$ is the reflection of $A$ about $(2,4)$.

Let the opposite vertex be $C=(h,k)$. Using midpoint formula, $\left(\frac{0+h}{2},\frac{3+k}{2}\right)=(2,4).$

So, $\frac{h}{2}=2 \implies h=4,$ $\frac{3+k}{2}=4 \implies 3+k=8 \implies k=5.$

Thus, $C=(4,5).$

Let the other two vertices be $B$ on one side through $A$ and $D$ on the other side through $A$.

Then the four vertices are $A,B,C,D$ with $A+C=B+D.$ Since $A+C=(0,3)+(4,5)=(4,8),$ we must have $B+D=(4,8).$

Now $B$ lies on one of the given lines and $D$ lies on the other.

Check the options to see which can be a vertex. If one option is $B$, then the corresponding other vertex is $D=(4,8)-B.$

• Option A: $B=(2,1)$ $2+1=3,$ so it lies on $x+y=3$. Then $D=(4,8)-(2,1)=(2,7).$ Check on other line: $2-7+3=-2\neq 0.$ Not possible.

• Option B: $B=(2,6)$ $2+6=8\neq 3,$ and $2-6+3=-1\neq 0.$ Not on either given side. Not possible.

• Option C: $B=(3,5)$ $3+5=8\neq 3,$ and $3-5+3=1\neq 0.$ Not on either given side. Not possible.

• Option D: $B=(3,6)$ Check: $3-6+3=0,$ so it lies on $x-y+3=0$. Then $D=(4,8)-(3,6)=(1,2).$ Check on the other line: $1+2=3,$ so it lies on $x+y=3$. Hence this is valid.

Therefore, one vertex of the parallelogram is $\boxed{(3,6)}.$