Let $$a - 2b + c = 1$$.
If $$f(x) = \begin{vmatrix} x+a & x+2 & x+1 \\ x+b & x+3 & x+2 \\ x+c & x+4 & x+3 \end{vmatrix}$$, then:

We begin with the 3 × 3 determinant

$$$f(x)=\begin{vmatrix} x+a & x+2 & x+1\\ x+b & x+3 & x+2\\ x+c & x+4 & x+3 \end{vmatrix},$$$

together with the given linear relation

$$a-2b+c = 1.$$

Our first aim is to simplify the determinant by elementary row operations, which do not change its value. We subtract the first row from the second and also from the third:

$$ R_2 \longleftarrow R_2-R_1,\qquad R_3 \longleftarrow R_3-R_1. $$

After these operations the determinant becomes

$$$ f(x)=\begin{vmatrix} x+a & x+2 & x+1\\ \;(x+b)-(x+a)&\;(x+3)-(x+2)&\;(x+2)-(x+1)\\ \;(x+c)-(x+a)&\;(x+4)-(x+2)&\;(x+3)-(x+1) \end{vmatrix} = \begin{vmatrix} x+a & x+2 & x+1\\ b-a & 1 & 1\\ c-a & 2 & 2 \end{vmatrix}. $$$

We next take advantage of the fact that the last two columns now show a clear pattern. To exploit it, we use a column operation that also leaves the determinant unchanged:

$$C_2 \longleftarrow C_2 - C_3.$$

This gives

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

Because the new second column now has zeros in its lower two entries, expansion along that column becomes very convenient. Remembering that the sign attached to the $$(1,2)$$ element is $$(-1)^{1+2}=-1$$, we write the expansion:

$$$ f(x)= -\,1 \times\begin{vmatrix} b-a & 1\\ c-a & 2 \end{vmatrix}. $$$

The 2 × 2 determinant inside is evaluated in the usual way:

$$$ \begin{vmatrix} b-a & 1\\ c-a & 2 \end{vmatrix} = (b-a)\cdot 2 - 1\cdot(c-a) = 2b-2a - c + a = 2b - a - c. $$$

Substituting this back, we obtain a remarkably simple expression for $$f(x)$$:

$$$ f(x)= -\,(2b - a - c)= a + c - 2b. $$$

Very importantly, notice that every $$x$$ has disappeared; the determinant is actually a constant, independent of $$x$$.

Next, we put the given condition $$a-2b+c=1$$ to use. Re-arranging that condition yields

$$ a + c - 2b = 1. $$

But this is exactly the expression we have just found for $$f(x)$$. Hence

$$ f(x)=1\quad\text{for every real }x. $$

In particular, when $$x=50$$ we have

$$ f(50)=1. $$

Therefore, among the options provided, the statement that matches our result is $$f(50)=1$$, which is Option D.

Hence, the correct answer is Option D.