If a, b, c are the lengths of the sides of a triangle 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 isosceles triangle

The correct answer is Option (2) → ΔABC is isosceles triangle

$Δ=\begin{vmatrix}a^2 & b^2 & c^2 \\(a+1)^2 & (b+1)^2 & (c+1)^2 \\(a-1)^2 & (b-1)^2 & (c-1)^2 \end{vmatrix}$

$R_2→R_2-R_3$

$Δ=4\begin{vmatrix}a^2 & b^2 & c^2 \\a & b & c \\(a-1)^2 & (b-1)^2 & (c-1)^2 \end{vmatrix}$

$R_3→R_3-R_1+2R_2$

$=4\begin{vmatrix}a^2 & b^2 & c^2 \\a & b & c \\1 & 1 & 1 \end{vmatrix}$

$=-4(a-b)(b-c)(c-a)=0$

$⇒a=b$ or $b=c$ or $c=a$

$a,b,c$ → isosceles