Question:

Step 1: Define the given function and condition. The function $f(x)$ is given as a $3 \times 3$ determinant: $$ f(x) = \begin{vmatrix} x+a & x+2 & x+1 \\ x+b & x+3 & x+2 \\ x+c & x+4 & x+3 \end{vmatrix} $$ We are also given the condition: $$ a - 2b + c = 1 $$ Step 2: Apply a row operation to simplify the determinant. To simplify the determinant, we apply the row operation $R_1 \rightarrow R_1 + R_3 - 2R_2$. This operation is chosen because it combines the terms $a, b, c$ in the desired form $a-2b+c$ and aims to create zeros in the first row. Step 3: Calculate the elements of the new first row. Let's apply the operation to each element of the first row: For the first element ($C_1$): $$(x+a) + (x+c) - 2(x+b) = x+a+x+c-2x-2b = a-2b+c$$ For the second element ($C_2$): $$(x+2) + (x+4) - 2(x+3) = x+2+x+4-2x-6 = 2x+6-2x-6 = 0$$ For the third element ($C_3$): $$(x+1) + (x+3) - 2(x+2) = x+1+x+3-2x-4 = 2x+4-2x-4 = 0$$ Step 4: Substitute the given condition into the simplified first row. Using the calculated new first row elements and the given condition $a - 2b + c = 1$, the first row of the determinant becomes: $$ \begin{vmatrix} 1 & 0 & 0 \\ x+b & x+3 & x+2 \\ x+c & x+4 & x+3 \end{vmatrix} $$ Step 5: Expand the determinant along the first row. Since the first row now contains two zeros, expanding the determinant along the first row simplifies the calculation significantly: $$ f(x) = 1 \cdot \begin{vmatrix} x+3 & x+2 \\ x+4 & x+3 \end{vmatrix} - 0 \cdot \begin{vmatrix} x+b & x+2 \\ x+c & x+3 \end{vmatrix} + 0 \cdot \begin{vmatrix} x+b & x+3 \\ x+c & x+4 \end{vmatrix} $$ $$ f(x) = 1 \cdot \begin{vmatrix} x+3 & x+2 \\ x+4 & x+3 \end{vmatrix} $$ Step 6: Evaluate the remaining $2 \times 2$ determinant. Now, we calculate the $2 \times 2$ determinant: $$ f(x) = (x+3)(x+3) - (x+2)(x+4) $$ $$ f(x) = (x^2 + 6x + 9) - (x^2 + 4x + 2x + 8) $$ $$ f(x) = (x^2 + 6x + 9) - (x^2 + 6x + 8) $$ $$ f(x) = x^2 + 6x + 9 - x^2 - 6x - 8 $$ $$ f(x) = 1 $$ Step 7: State the final result for $f(x)$ and calculate $f(50)$. We found that $f(x) = 1$ for all values of $x$. Therefore, to find $f(50)$, we substitute $x=50$ into the expression for $f(x)$: $$ f(50) = 1 $$ The final answer is $f(50)=1$, which corresponds to Option 3. The final answer is $\boxed{\text{f(50) = 1}}$.