Answer
0
Work Step by Step
Let \[A=\left|\begin{array}{ccc} a&b&c\\
a+x&b+x&c+x\\
a+y&b+y&c+y\end{array}\right|\]
Using column operation $C_1\rightarrow C_1-C_2$
\[=\left|\begin{array}{ccc} a-b&b&c\\
a-b&b+x&c+x\\
a-b&b+y&c+y\end{array}\right|\]
We will use the property of determinant
\[\left|\begin{array}{ccc}k p&q&r\\
ks&t&u\\
kv&w&\alpha\end{array}\right|=k\left|\begin{array}{ccc} p&q&r\\
s&t&u\\
v&w&\alpha\end{array}\right|\;\;\;...(1)\]
Using (1)
\[=(a-b)\left|\begin{array}{ccc} 1&b&c\\
1&b+x&c+x\\
1&b+y&c+y\end{array}\right|\]
Applying row operation $R_2\rightarrow R_2-R_1$ and $R_3\rightarrow R_3-R_1$
\[=(a-b)\left|\begin{array}{ccc} 1&b&c\\
0&x&x\\
0&y&y\end{array}\right|\]
Using (1) in row $R_2$ and $R_3$
\[=(a-b)xy\left|\begin{array}{ccc} 1&b&c\\
0&1&1\\
0&1&1\end{array}\right|\]
Now $R_2$ and $R_3$ are identical so value of determinant on R.H.S. will be zero
\[(a-b)xy\left|\begin{array}{ccc} 1&b&c\\
0&1&1\\
0&1&1\end{array}\right|=(a-b)xy(0)=0\]
Hence $A=0$.
