Let us consider first even natural number be $$ 2x$$
Second even number be $$ 2 x+2$$
Third even number be $$2 x+4$$
So according to the question,
$$(2 x)^{2}+(2 x+2)^{2}+(2 x+4)^{2}=308$$
$$4 x^{2}+4 x^{2}+8 x+4+4 x^{2}+16 x+16-308=0$$
$$12 x^{2}+24 x-288=0$$
Divide by $$12,$$ we get $$x^{2}+2 x-24=0$$
Let us factorize,
$$x^{2}+6 x-4 x-24=0$$
$$x(x+6)-4(x+6)=0$$
$$(x+6)(x-4)=0$$
$$\mathrm{So}$$
$$(x+6)=0$$ or $$(x-4)=0$$
$$x=-6$$ or $$x=4$$
$$\therefore$$ Value of $$x=4$$ $$[\because x=-6$$ is not an even natural number $$]$$
First even natural number $$=2 x=2(4)=8$$
Second even natural number
$$=2 x+2=2(4)+2=10$$
Third even natural number
$$=2 x+4=2(4)+4=12$$
$$\therefore$$ The numbers are $$ 8,~10,$$ and $$12$$