Answer
See the proof below.
Work Step by Step
We have:
$$
\left|\begin{array}{ccc}
1+a & 1 & 1 \\
1 & 1+b & 1 \\
1 & 1 & 1+c
\end{array}\right|=a b c \frac{111}{a b c}\left|\begin{array}{ccc}
1+a & 1 & 1 \\
1 & 1+b & 1 \\
1 & 1+c
\end{array}\right|=a b c\left|\begin{array}{ccc}
\frac{1}{a}+1 & \frac{1}{b} & \frac{1}{c} \\
\frac{1}{a} & \frac{1}{b}+1 & \frac{1}{c} \\
\frac{1}{a} & \frac{1}{b} & \frac{1}{c}+1
\end{array}\right|
$$
Adding the second column and the third column to the first column, we have
$$
\left|\begin{array}{ccc}
1+a & 1 & 1 \\
1 & 1+b & 1 \\
1 & 1 & 1+c
\end{array}\right|=a b c\left|\begin{array}{ccc}
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1 & \frac{1}{b} & \frac{1}{c} \\
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1 & \frac{1}{b}+1 & \frac{1}{c} \\
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1 & \frac{1}{b} & \frac{1}{c}+1
\end{array}\right|=a b c\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\right)\left|\begin{array}{ccc}
1 & \frac{1}{b} & \frac{1}{c} \\
1 & \frac{1}{b}+1 & \frac{1}{c} \\
1 & \frac{1}{b} & \frac{1}{c}+1
\end{array}\right|
$$
Multiply the first row with (-1) and add to the second row. Then, multiply the first row with (-1) and add to the third row.
$$
\left|\begin{array}{ccc}
1+a & 1 & 1 \\
1 & 1+b & 1 \\
1 & 1 & 1+c
\end{array}\right|=a b c\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\right)\left|\begin{array}{cc}
1 & \frac{1}{b} & \frac{1}{c} \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right|=a b c\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\right)
$$
