If a^2+b^2+c^2=-2 and f(x)= | … | then f(x) is a polynomial of degree

$$\begin{aligned} & f(x)=\left|\begin{array}{ccc} 1+\left(a^2+b^2+c^2+2\right) x & \left(1+b^2\right) x & \left(1+c^2\right) x \\ 1+\left(a^2+b^2+c^2+2\right) x & 1+b^2 x & \left(1+c^2\right) x \\ 1+\left(a^2+b^2+c^2+2\right) x & \left(1+b^2\right) x & 1+c^2 x \end{array}\right| \text {, Applying } C_1 \rightarrow C_1+C_2+C_3 \\ & =\left|\begin{array}{ccc} 1 & \left(1+b^2\right) x & \left(1+c^2\right) x \\ 1 & 1+b^2 x & \left(1+c^2\right) x \\ 1 & \left(1+b^2\right) x & 1+c^2 x \end{array}\right| \because a^2+b^2+c^2+2=0 \\ & f(x)=\left|\begin{array}{ccc} 0 & x-1 & 0 \\ 0 & 1-x & x-1 \\ 1 & \left(1+b^2\right) x & 1+c x \end{array}\right| ; \text { Applying } R_1 \rightarrow R_1-R_2, R_2 \rightarrow R_2-R_3 \\ & f(x)=(x-1)^2 \end{aligned}$$​f(x)=​1+(a2+b2+c2+2)x1+(a2+b2+c2+2)x1+(a2+b2+c2+2)x​(1+b2)x1+b2x(1+b2)x​(1+c2)x(1+c2)x1+c2x​​, Applying C1​→C1​+C2​+C3​=​111​(1+b2)x1+b2x(1+b2)x​(1+c2)x(1+c2)x1+c2x​​∵a2+b2+c2+2=0f(x)=​001​x−11−x(1+b2)x​0x−11+cx​​; Applying R1​→R1​−R2​,R2​→R2​−R3​f(x)=(x−1)2​ Hence degree $=2$=2.