1. Find the pairwise intersections of the three lines

The lines are: $L_1: x-y=0 \implies y=x$ $L_2: x+2y=3$ $L_3: 2x+y=6$

These three lines intersect pairwise to form a triangle. Let us find its vertices.

2. Intersection of $L_1$ and $L_2$

From $L_1$, $y=x$. Substitute into $L_2$: $x+2x=3$ $3x=3 \implies x=1$ So $y=1$.

Hence, $A=(1,1)$

3. Intersection of $L_1$ and $L_3$

From $L_1$, $y=x$. Substitute into $L_3$: $2x+x=6$ $3x=6 \implies x=2$ So $y=2$.

Hence, $B=(2,2)$

4. Intersection of $L_2$ and $L_3$

Solve $x+2y=3$ $2x+y=6$

From the first equation, $x=3-2y$ Substitute into the second: $2(3-2y)+y=6$ $6-4y+y=6$ $-3y=0 \implies y=0$ Then, $x=3$

Hence, $C=(3,0)$

5. Find the side lengths of triangle $ABC$

Using distance formula:

$AB=\sqrt{(2-1)^2+(2-1)^2}=\sqrt{1+1}=\sqrt{2}$

$BC=\sqrt{(3-2)^2+(0-2)^2}=\sqrt{1+4}=\sqrt{5}$

$CA=\sqrt{(3-1)^2+(0-1)^2}=\sqrt{4+1}=\sqrt{5}$

So, $BC=CA=\sqrt{5}$

Hence the triangle has two equal sides, so it is isosceles.

6. Check if it is right-angled

For a right triangle, Pythagoras must hold.

Largest side is $\sqrt{5}$. Check: $AB^2+BC^2=2+5=7 \neq 5$ $AB^2+CA^2=2+5=7 \neq 5$ $BC^2+CA^2=5+5=10 \neq 2$

So it is not right-angled.

7. Check if it is equilateral

Side lengths are: $\sqrt{2},\ \sqrt{5},\ \sqrt{5}$ These are not all equal, so it is not equilateral.

8. Conclusion

The triangle formed is an isosceles triangle.

Therefore, the correct option is: $\boxed{\text{D: Isosceles triangle}}$

The derived answer matches the stored correct answer.