If a, b, & c are the lengths of the sides of a ΔABC and $\begin{vmatrix}a^2&b^2&c^2\\(a+1)^2&(b+1)^2&(c+1)^2\\(a-1)^2&(b-1)^2&(c-1)^2\end{vmatrix}= 0$, then

ΔABC is an Isosceles triangle

When a = b or b = c or c = a the determinant reduces to zero.

It is not necessary that a = b = c for the determinant to be zero.

Therefore the triangle is isosceles.

Hence (C) is the correct answer.