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

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

Step 1: Row Operations

Subtract the first row from the second row and the third row:
$$R_2 \to R_2 - R_1$$
$$R_3 \to R_3 - R_1$$ 

Performing these operations yields:

Row 2 components:

$$(1+a^2)x - (1+a^2x) = x + a^2x - 1 - a^2x = x - 1$$

$$1+b^2x - (1+b^2)x = 1 + b^2x - x - b^2x = 1 - x$$

$$(1+c^2)x - (1+c^2)x = 0$$ 

Row 3 components:

$$(1+a^2)x - (1+a^2x) = x - 1$$

$$(1+b^2)x - (1+b^2)x = 0$$

$$1+c^2x - (1+c^2)x = 1 - x$$ 

Now, our determinant becomes:

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

Step 2: Column Operation

Add the second column and third column to the first column:

$$C_1 \to C_1 + C_2 + C_3$$ Let us compute the new entries for the first column:

Row 1, Column 1:

$$(1+a^2x) + (1+b^2)x + (1+c^2)x = 1 + a^2x + x + b^2x + x + c^2x$$

$$= 1 + 2x + (a^2+b^2+c^2)x$$

Given that $$a^2+b^2+c^2 = -2$$ , substituting this value results in:
$$1 + 2x + (-2)x = 1$$ 

Row 2, Column 1:
$$(x-1) + (1-x) + 0 = 0$$ 

Row 3, Column 1:
$$(x-1) + 0 + (1-x) = 0$$

Thus, the simplified determinant transforms into a lower triangular shape:
$$f(x) = \begin{vmatrix} 1 & (1+b^2)x & (1+c^2)x \\ 0 & 1-x & 0 \\ 0 & 0 & 1-x \end{vmatrix}$$ 

Step 3: Expansion
Expanding along the first column gives:

$$f(x) = 1 \cdot \begin{vmatrix} 1-x & 0 \\ 0 & 1-x \end{vmatrix}$$

$$f(x) = (1-x)(1-x) - 0 = (1-x)^2$$

$$f(x) = x^2 - 2x + 1$$

Therefore the degree of the ploynomial $$f(x)$$ is 2.