If $a^2 +b^2+c^3 + ab + bc + ca ≤0$ for all $a, b, c ∈ R$, then the value of the determinant $\begin{vmatrix}(a+b+2)^2&a^2+b^2&1\\1&(b+c+2)^2& b^2+c^2\\c^2 + a^2&1(c+a+2)^2\end{vmatrix}$, is equal to

65

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 $a+b+c=0$.

$⇒a=b=c=0$

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