If (2a - 3)$$^{2}$$ + (3b + 4)$$^{2}$$ + (6c + 1)$$^{2}$$ = 0, then value of $$\frac{a^{3}+b^{3}+c^{3}-3abc}{a^{2}+b^{2}+c^{2}}+3$$ is:

Given : $$(2a-3)^2+(3b+4)^2+(6c+1)^2=0$$

Sum of three positive terms is zero, iff all the three terms are zero.

=> $$2a-3=0$$

=> $$a=\frac{3}{2}$$

Similarly, $$b=\frac{-4}{3}$$ and $$c=\frac{-1}{6}$$

Now, $$a+b+c=\frac{3}{2}-\frac{4}{3}-\frac{1}{6}=0$$ -----------(i)

Also, $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$$

Substituting value from equation (i), we get :

=> $$a^3+b^3+c^3-3abc=0$$ ------------(ii)

$$\therefore$$ $$\ \frac{a^{3}+b^{3}+c^{3}-3abc}{a^{2}+b^{2}+c^{2}}\ =0$$

=> Ans - (C)