The value of $Δ=\begin{bmatrix}1 & a & b+c\\1 & b & c+a\\1 & c & a+b\end{bmatrix},$ is

1

The correct answer is option (4) : 0

Applying $C_2→C_2+C_3,$ we get

$Δ=\begin{bmatrix}1 & a+b+c & b+c\\1 & b+c+a & c+a\\1 & c+a+b & a+b\end{bmatrix}$

$⇒Δ= (a+b+c) \begin{bmatrix}1 & 1 & b+c\\1 & 1 & c+a\\1 & 1 & a+b\end{bmatrix}$           $[ \begin{bmatrix}\text{Taking out a+b + c}\\\text{common from} C_2\end{bmatrix}$    

$⇒Δ=(a+b+c)×0=0$                  $[∵ C_1 \, and \, C_2\, are \, identical]$