Determinant of a Triangular Matrix

Column operation $$C_1 \to C_1+C_2+C_3$$

All rows of new $$C_1$$ equal $$x+6$$, so:

$$f(x) = (x+6)\begin{vmatrix} 1 & 2 & 3 \\ 1 & x+2 & 3 \\ 1 & 2 & x+3 \end{vmatrix}$$

Row operations $$R_2 \to R_2-R_1$$, $$R_3 \to R_3-R_1$$

$$f(x) = (x+6)\begin{vmatrix} 1 & 2 & 3 \\ 0 & x & 0 \\ 0 & 0 & x \end{vmatrix} = x^2(x+6)$$

$$\alpha = 0,\quad \beta = 0,\quad \gamma = -6$$

$$\alpha^3+\beta^3+\gamma^3 - 3\alpha\beta\gamma = (\alpha+\beta+\gamma)\left[(\alpha+\beta+\gamma)^2 - 3(\alpha\beta+\beta\gamma+\gamma\alpha)\right]$$

$$\alpha+\beta+\gamma = -6$$, $$;\alpha\beta+\beta\gamma+\gamma\alpha = 0$$, $$;\alpha\beta\gamma = 0$$:

$$= (-6)\left[(-6)^2 - 3(0)\right] = (-6)(36) = \boxed{-216}$$

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.