If ${a^2} + {b^2} + {c^2} = - 2$ and $f(x) = \left| {\begin{array}{*{20}{l}}
  {1 + {a^2}x}&{\left( {1 + {b^2}} \right)x}&{\left( {1 + {c^2}} \right)x} \\
  {\left( {1 + {a^2}} \right)x}&{1 + {b^2}x}&{\left( {1 + {c^2}} \right)x} \\
  {\left( {1 + {a^2}} \right)x}&{\left( {1 + {b^2}} \right)x}&{1 + {c^2}x}
\end{array}} \right|$. Then, $f(x)$ is a polynomial of degree
(A)$2$
(B)$3$
(C)$4$
(D)$5$

Hint:
The degree of a polynomial is the highest power of the variable in a polynomial
expression. For ex- The degree of a polynomial $6{x^4} + 2{x^3} + 1$ is $4$. In the given
question, we must solve $f\left( x \right)$ by using the properties of determinants to find its degree.

Complete step by step solution:
Given, $f(x) = \left| {\begin{array}{*{20}{l}}
  {1 + {a^2}x}&{\left( {1 + {b^2}} \right)x}&{\left( {1 + {c^2}} \right)x} \\
  {\left( {1 + {a^2}} \right)x}&{1 + {b^2}x}&{\left( {1 + {c^2}} \right)x} \\
  {\left( {1 + {a^2}} \right)x}&{\left( {1 + {b^2}} \right)x}&{1 + {c^2}x}
\end{array}} \right|$
Solve the $f(x)$ by using elementary row and column transformations.
Applying ${C_1} \to {C_1} + {C_2} + {C_3}$ to $f(x)$, we get
$f(x) = \left| {\begin{array}{*{20}{l}}
  {1 + {a^2}x + \left( {1 + {b^2}} \right)x + \left( {1 + {c^2}} \right)x}&{\left( {1 + {b^2}} \right)x}&{\left( {1 + {c^2}} \right)x} \\
  {\left( {1 + {a^2}} \right)x + 1 + {b^2}x + \left( {1 + {c^2}} \right)x}&{1 + {b^2}x}&{\left( {1 + {c^2}} \right)x} \\
  {\left( {1 + {a^2}} \right)x + \left( {1 + {b^2}} \right)x + 1 + {c^2}x}&{\left( {1 + {b^2}} \right)x}&{1 + {c^2}x}
\end{array}} \right|$
$ \Rightarrow $$f(x) = \left| {\begin{array}{*{20}{l}}
  {1 + {a^2}x + x + {b^2}x + x + {c^2}x}&{\left( {1 + {b^2}} \right)x}&{\left( {1 + {c^2}} \right)x} \\
  {x + {a^2}x + 1 + {b^2}x + x + {c^2}x}&{1 + {b^2}x}&{\left( {1 + {c^2}} \right)x} \\
  {x + {a^2}x + x + {b^2}x + 1 + {c^2}x}&{\left( {1 + {b^2}} \right)x}&{1 + {c^2}x}
\end{array}} \right|$
$ \Rightarrow $$f(x) = \left| {\begin{array}{*{20}{l}}
  {1 + 2x + \left( {{a^2} + {b^2} + {c^2}} \right)x}&{\left( {1 + {b^2}} \right)x}&{\left( {1 + {c^2}} \right)x} \\
  {1 + 2x + \left( {{a^2} + {b^2} + {c^2}} \right)x}&{1 + {b^2}x}&{\left( {1 + {c^2}} \right)x} \\
  {1 + 2x + \left( {{a^2} + {b^2} + {c^2}} \right)x}&{\left( {1 + {b^2}} \right)x}&{1 + {c^2}x}
\end{array}} \right|$
Now put ${a^2} + {b^2} + {c^2} = - 2$ (Given)
$\therefore $$f(x) = \left| {\begin{array}{*{20}{l}}
  {1 + 2x + \left( { - 2} \right)x}&{\left( {1 + {b^2}} \right)x}&{\left( {1 + {c^2}} \right)x} \\
  {1 + 2x + \left( { - 2} \right)x}&{1 + {b^2}x}&{\left( {1 + {c^2}} \right)x} \\
  {1 + 2x + \left( { - 2} \right)x}&{\left( {1 + {b^2}} \right)x}&{1 + {c^2}x}
\end{array}} \right|$
$ \Rightarrow $$f(x) = \left| {\begin{array}{*{20}{l}}
  {1 + 2x - 2x}&{\left( {1 + {b^2}} \right)x}&{\left( {1 + {c^2}} \right)x} \\
  {1 + 2x - 2x}&{1 + {b^2}x}&{\left( {1 + {c^2}} \right)x} \\
  {1 + 2x - 2x}&{\left( {1 + {b^2}} \right)x}&{1 + {c^2}x}
\end{array}} \right|$
$ \Rightarrow $$f(x) = \left| {\begin{array}{*{20}{l}}
  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|$
Now applying ${R_2} \to {R_2} - {R_1}$ and${R_3} \to {R_3} - {R_1}$ to $f(x)$, we get
$f(x) = \left| {\begin{array}{*{20}{l}}
  1&{\left( {1 + {b^2}} \right)x}&{\left( {1 + {c^2}} \right)x} \\
  {1 - 1}&{1 + {b^2}x - \left( {1 + {b^2}} \right)x}&{\left( {1 + {c^2}} \right)x - \left( {1 + {c^2}} \right)x} \\
  {1 - 1}&{\left( {1 + {b^2}} \right)x - \left( {1 + {b^2}} \right)x}&{1 + {c^2}x - \left( {1 + {c^2}} \right)x}
\end{array}} \right|$
$ \Rightarrow f(x) = \left| {\begin{array}{*{20}{l}}
  1&{\left( {1 + {b^2}} \right)x}&{\left( {1 + {c^2}} \right)x} \\
  0&{1 + {b^2}x - x - {b^2}x}&0 \\
  0&0&{1 + {c^2}x - x - {c^2}x}
\end{array}} \right|$
$ \Rightarrow f(x) = \left| {\begin{array}{*{20}{l}}
  1&{\left( {1 + {b^2}} \right)x}&{\left( {1 + {c^2}} \right)x} \\
  0&{1 - x}&0 \\
  0&0&{1 - x}
\end{array}} \right|$
Now expanding the determinant $f(x)$ along ${C_1}$, we obtain
$ \Rightarrow f(x) = 1\left\{ {\left( {1 - x} \right)\left( {1 - x} \right) - \left( {0 \times 0} \right)} \right\}$
$ \Rightarrow f(x) = {\left( {1 - x} \right)^2}$
$ \Rightarrow f(x) = 1 + {x^2} - 2x$
As we know that the degree of a polynomial is the highest power of the variable in a polynomial expression. Here the highest power of $x$ is $2$.
Therefore, the degree of the polynomial $f\left( x \right)$ is $2$.

Hence, option (A) is the correct answer.

Note:
While solving the determinant , try to make three $1's$ in any row or column, so that determinant can be solved easily. Also, you must have the knowledge to apply the properties of determinant while solving it.