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

$f(50) = 1$

The correct answer is option (1) : $f(50) = 1$

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

$⇒f(x)=\begin{vmatrix}a-2b+c & 0 & 0\\x+b & x+3 & x+2\\x+c & x+4 & x+3\end{vmatrix}$

Applying $R_1→R_1-2R_2+R_3$

$⇒f(x)=\begin{vmatrix}1 & 0 & 0\\x+b & x+3 & x+2\\x+c & x+4 & x+3\end{vmatrix}= 1\, ∀ \, x$

$⇒f(50)=1$