If $a^2+b^2 +c^3 + ab + bc + ca ≤ 0 \, ∀ a, b, c, \in R, $ then the value of the determinant

$\begin{vmatrix}(a+b+c)^2 & a^2+b^2 & 1\\1 & (b+c+2)^2 & b^2+c^2\\c^2+a^2 & 1 & (c+a+2)^2\end{vmatrix}, $ is equal to

65

The correct answer is option (1) : 65

We have,

$a^2+b^2 +c^3+ab +bc + ca ≤0$

$⇒2a^2+2b^2+2c^2+2ab + 2bc + 2ca ≤0$

$⇒(a+b)^2 (b+c)^2 +(c+a)^2 ≤0$

$⇒a+ b =0, b + c =0, c+a=0$

$⇒(a+b) +(b+c) + (c+a) = 0 $

$⇒ a+b + c= 0 $

Thus, we have $a+b = 0, b+c =0, c+a=0 $ and $a+b + c= 0.$

$⇒a=b=c=0$

$∴\begin{bmatrix}(a+b+c)^2&a^2+b^2& 1\\1&(b+c+2)^2 & b^2+c^2\\c^2+a^2&1&(c+a+2)^2\end{bmatrix}= \begin{bmatrix} 4&0&1\\1&4&0\\0&1&4\end{bmatrix}= 65 $