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

$(a)\;1\qquad(b)\;0\qquad(c)\;3\qquad(d)\;2$

$f(x)=\begin{vmatrix}1+a^2x&(1+b^2)x&(1+c^2)x\\(1+a^2)x&1+b^2x&(1+c^2)x\\(1+a^2)x&(1+b^2)x&1+c^2x\end{vmatrix}$
Apply $C_1\rightarrow C_1+C_2+C_3$ we get,
$F(x)=\begin{vmatrix}1+(a^2+b^2+c^2+2)x&(1+b^2)x&(1+c^2)x\\1+(a^2+b^2+c^2+2)x&1+b^2x&(1+c^2x)\\1+(a^2+b^2+c^2+2)x&(1+b^2)x&1+c^2x\end{vmatrix}$
$\;\;\;\;\;\;=\begin{vmatrix}1&(1+b^2)x&(1+c^2)x\\1&1+b^2x&(1+c^2x)\\1&(1+b^2)x&1+c^2x\end{vmatrix}$
As given that $a^2+b^2+c^2=-2$
$\therefore a^2+b^2+c^2+2=0$
Apply $R_1\rightarrow R_1-R_2,R_2\rightarrow R_2-R_3$
$\therefore f(x)=\begin{vmatrix}0 &x-1&0\\0&1-x&x-1\\1&(1+b^2)x&1+c^2x\end{vmatrix}$
$f(x)=(x-1)^2$ Hence degree =2
Hence (d) is the correct answer.