If the vectors \(\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+a \hat{\mathbf{j}}+a^{2} \hat{\mathbf{k}}\),
\(\overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}+b \hat{\mathbf{j}}+b^{2} \hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{c}}=\hat{\mathbf{i}}+c \hat{\mathbf{j}}+c^{2} \hat{\mathbf{k}}\) are three
non-coplanar vectors and \(\left|\begin{array}{lll}a & a^{2} & 1+a^{3} \\ b & b^{2} & 1+b^{3} \\ c & c^{2} & 1+c^{3}\end{array}\right|=0\),
then the value of \(a b c\) is

Since, \(\overrightarrow{ a }, \overrightarrow{ b }\), and \(\overrightarrow{ c }\) are non-coplanar vectors,
therefore \([\overrightarrow{ a } \overrightarrow{ b } \overrightarrow{ c }] \neq 0\)
\(\Rightarrow \Delta=\left|\begin{array}{lll}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\right| \neq 0\)
\(\Rightarrow\) \(\Delta \neq 0\)
Now, \(\left(\begin{array}{lll}a & a^2 & 1+a^3 \\ b & b^2 & 1+b^3 \quad \mid=0 \\ c & c^2 & 1+c^3\end{array}\right.\)
\(\Rightarrow\left|\begin{array}{lll}a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1\end{array}\right|+\left|\begin{array}{lll}a & a^2 & a^3 \\ b & b^2 & b^3 \\ c & c^2 & c^3\end{array}\right|=0\)
\(\Rightarrow\) \(\Delta(1+a b c)=0\)
\(\Rightarrow\) \(a b c=-1\)