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(C) Given $a^2 + b^2 + c^2 = -2$. Apply the column operation $C_1 \rightarrow C_1 + C_2 + C_3$: $f(x) = \left| \begin{array}{ccc} 1 + x(a^2 + b^2 + c^

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(C) Given $a^2 + b^2 + c^2 = -2$. Apply the column operation $C_1 ightarrow C_1 + C_2 + C_3$: $f(x) = \left| \begin{array}{ccc} 1 + x(a^2 + b^2 + c^2 + 2) & (1 + b^2)x & (1 + c^2)x \ 1 + x(a^2 + b^2 + c^2 + 2) & 1 + b^2x & (1 + c^2)x \ 1 + x(a^2 + b^2 + c^2 + 2) & (1 + b^2)x & 1 + c^2x \end{array} ight|$ Since $a^2 + b^2 + c^2 = -2$,the term $a^2 + b^2 + c^2 + 2 = 0$. Thus,$f(x) = \left| \begin{array}{ccc} 1 & (1 + b^2)x & (1 + c^2)x \ 1 & 1 + b^2x & (1 + c^2)x \ 1 & (1 + b^2)x & 1 + c^2x \end{array} ight|$ Apply row operations $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_1$: $f(x) = \left| \begin{array}{ccc} 1 & (1 + b^2)x & (1 + c^2)x \ 0 & 1 - x & 0 \ 0 & 0 & 1 - x \end{array} ight|$ Expanding along the first column: $f(x) = 1 \cdot (1 - x)(1 - x) = (1 - x)^2 = 1 - 2x + x^2$. Since the highest power of $x$ is $2$,$f(x)$ is a polynomial of degree $2$.

If $a^2 + b^2 + c^2 = -2$ and $f(x) = \left| \begin{array}{ccc} 1 + a^2x & (1 + b^2)x & (1 + c^2)x \\ (1 + a^2)x & 1 + b^2x & (1 + c^2)x \\ (1 + a^2)x & (1 + b^2)x & 1 + c^2x \end{array} \right|$,then $f(x)$ is a polynomial of degree

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