Answer
0
Work Step by Step
Let \[A=\left|\begin{array}{ccc} 1&a&b+c\\
1&b&a+c\\
1&c&a+b\end{array}\right|\]
Applying column operation on $A$
\[C_3\rightarrow C_3+C_2\]
\[=\left|\begin{array}{ccc} 1&a&a+b+c\\
1&b&a+b+c\\
1&c&a+b+c\end{array}\right|\]
We will use the property of determinant
\[\left|\begin{array}{ccc} x_1&y_1&kz_1\\
x_2&y_2&kz_2\\
x_3&y_3&kz_3\end{array}\right|=k\left|\begin{array}{ccc} x_1&y_1&z_1\\
x_2&y_2&z_2\\
x_3&y_3&z_3\end{array}\right|\;\;\;...(1)\]
Using (1)
\[\left|\begin{array}{ccc} 1&a&a+b+c\\
1&b&a+b+c\\
1&c&a+b+c\end{array}\right|=(a+b+c)\left|\begin{array}{ccc} 1&a&1\\
1&b&1\\
1&c&1\end{array}\right|\]
Now $C_1$ and $C_3$ are identical so value of determinant on R.H.S. will be zero
\[\Rightarrow(a+b+c)\left|\begin{array}{ccc} 1&a&1\\
1&b&1\\
1&c&1\end{array}\right|=(a+b+c)(0)=0\]
Hence $A=0$.
