To evaluate the determinant without directly expanding it, we can use standard determinant row and column operations to simplify the matrix.
Let the given determinant be $$\Delta$$:
$$\Delta = \begin{vmatrix}a^2+1 & ab & ac \\ab & b^2+1 & bc \\ac & bc & c^2+1\end{vmatrix}$$
Step 1: Multiply rows and divide the determinant
Multiply Row 1 ($$R_1$$) by $$a$$, Row 2 ($$R_2$$) by $$b$$, and Row 3 ($$R_3$$) by $$c$$. To keep the value unchanged, we divide the entire determinant by $$abc$$:
$$\Delta = \frac{1}{abc} \begin{vmatrix}a(a^2+1) & a^2b & a^2c \\ab^2 & b(b^2+1) & b^2c \\ac^2 & bc^2 & c(c^2+1)\end{vmatrix}$$
Step 2: Take out common factors from columns
Now, take out the common factor $$a$$ from Column 1 ($$C_1$$), $$b$$ from Column 2 ($$C_2$$), and $$c$$ from Column 3 ($$C_3$$):
$$\Delta = \frac{abc}{abc} \begin{vmatrix}a^2+1 & a^2 & a^2 \\b^2 & b^2+1 & b^2 \\c^2 & c^2 & c^2+1\end{vmatrix}$$
This simplifies to:
$$\Delta = \begin{vmatrix}a^2+1 & a^2 & a^2 \\b^2 & b^2+1 & b^2 \\c^2 & c^2 & c^2+1\end{vmatrix}$$
Step 3: Apply Row Operations
Add Row 2 and Row 3 to Row 1 ($$R_1 \rightarrow R_1 + R_2 + R_3$$):
$$\Delta = \begin{vmatrix}1+a^2+b^2+c^2 & 1+a^2+b^2+c^2 & 1+a^2+b^2+c^2 \\b^2 & b^2+1 & b^2 \\c^2 & c^2 & c^2+1\end{vmatrix}$$
Step 4: Factor out the common term
Take out $$(1 + a^2 + b^2 + c^2)$$ from the first row:
$$\Delta = (1+a^2+b^2+c^2) \begin{vmatrix}1 & 1 & 1 \\b^2 & b^2+1 & b^2 \\c^2 & c^2 & c^2+1\end{vmatrix}$$
Step 5: Perform Column Operations to create zeros
Apply $$C_2 \rightarrow C_2 - C_1$$ and $$C_3 \rightarrow C_3 - C_1$$:
$$\Delta = (1+a^2+b^2+c^2) \begin{vmatrix}1 & 0 & 0 \\b^2 & 1 & 0 \\c^2 & 0 & 1\end{vmatrix}$$
Since the remaining matrix is an upper/lower triangular matrix configuration, the determinant of this simplified matrix is simply the product of its diagonal elements ($$1 \times 1 \times 1 = 1$$).
$$\Delta = (1+a^2+b^2+c^2) \cdot 1 = 1 + a^2 + b^2 + c^2$$
Thus, the final value matches Option C.
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