We have,
$$
\Delta=\left|\begin{array}{ccc}
1 & 1 & 1 \\
1+\cos \mathrm{A} & 1+\cos \mathrm{B} & 1+\cos \mathrm{C} \\
\cos ^2 \mathrm{~A}+\cos \mathrm{A} & \cos ^2 \mathrm{~B}+\cos \mathrm{B} & \cos ^2 \mathrm{C}+\cos \mathrm{C}
\end{array}\right|=0
$$



$$
\begin{aligned}
&\Delta=\left|\begin{array}{ccc}
0 & 0 & 1 \\
\cos \mathrm{A}-\cos \mathrm{C} & \cos \mathrm{B}-\cos \mathrm{C} & 1+\cos \mathrm{C} \\
\cos ^2 \mathrm{~A}+\cos \mathrm{A}-\cos ^2 \mathrm{C}-\cos \mathrm{C} & \cos ^2 \mathrm{~B}+\cos \mathrm{B}-\cos ^2 \mathrm{C}-\cos & \mathrm{C} \cos ^2 \mathrm{C}+\cos \mathrm{C}
\end{array}\right|=0\\
&\left.\Rightarrow(\cos A-\cos C) \cdot(\cos B-\cos C) \quad C_1-C_3 \text { and } C_2 \rightarrow C_2-C_3\right]\\
&\left|\begin{array}{ccc}
0 & 0 & 1 \\
1 & 1 & 1+\cos C \\
\cos A+\cos C+1 & \cos B+\cos C+1 & \cos ^2 C+\cos C
\end{array}\right|=0\\
&\Rightarrow(\cos \mathrm{A}-\cos \mathrm{C}) \cdot(\cos \mathrm{B}-\cos \mathrm{C})\\
&[(\cos \mathrm{B}+\cos \mathrm{C}+1)-(\cos \mathrm{A}+\cos \mathrm{C}+1)]=0\\
&\Rightarrow(\cos A-\cos C) \times(\cos B-\cos C)\\
&(\cos B-\cos A)=0\\
&\text { i.e., } \cos A=\cos C \text { or } \cos B=\cos C\\
&\text { or } \cos \mathrm{B}=\cos \mathrm{A}\\
&\Rightarrow \mathrm{A}=\mathrm{C} \text { or } \mathrm{B}=\mathrm{C} \text { or } \mathrm{B}=\mathrm{A}\\
&\text { Hence, ABC is an isosceles triangle. }
\end{aligned}
$$