If $a^2 +b^2+c^2 = -2 $ and

$f(x)=\begin{vmatrix}1+a^2x & (1+b^2)x & (1+c^2)x\\(1+a^2)x & 1+b^2x & (1+c^2)x\\(1+a^2)x & (1+b^2)x & 1+c^2x\end{vmatrix}$, then f(x) is a polynomial of degree

2

The correct answer is option (1) : 2

Applying $C_1→C_1+C_2+C_3,$ we have

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

$⇒f(x) =\begin{vmatrix}1 & (1+b^2)x & (1+c^2)x\\1 & 1+b^2x & (1+c^2)x\\1 & (1+b^2)x & 1+c^2x\end{vmatrix}$     $[∵ a^2+b^2 +c^2 + 2= 0 ]$

$⇒f(x) =\begin{vmatrix}1 & (1+b^2)x & (1+c^2)x\\0 & 1-x &0\\0 & 0 & 1-x\end{vmatrix}$    $\begin{bmatrix}Applying\,  R_2→R_2-R_1\\R_3→R_3-R_1\end{bmatrix}$

$⇒f(x) = (1-x)^2,$ which is a polynomial of degree 2.