Theorem of Coplanar Vectors

If $\left|\begin{array}{ccc}a& a^2& 1+a^3\\ b& b^2& 1+b^3\\ c& c^2& 1+c^3\end{array}\right|=0$ and vectors$\left(1,a,a^2\right)$, $\left(1,b,b^2\right)$$\left(1,c,c^2\right)$ are non-coplanar, then the product $abc$ is equal to:

If $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ are non-coplanar vectors, then prove that the four points $2\overrightarrow{a}+3\overrightarrow{b}–\overrightarrow{c}$, $\overrightarrow{a}–2\overrightarrow{b}+3\overrightarrow{c}$, $3\overrightarrow{a}+4\overrightarrow{b}–2\overrightarrow{c}$ and $\overrightarrow{a}–6\overrightarrow{b}+6\overrightarrow{c}$ are coplanar

Show that the vectors $\left(\overrightarrow{a}-2\overrightarrow{b}+3\overrightarrow{c}\right),\left(-2\overrightarrow{a}+3\overrightarrow{b}-4\overrightarrow{c}\right) \&\left(-\overrightarrow{b}+2\overrightarrow{c}\right)$ are coplanar vector, where $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are non-coplanar vectors.

If the vectors $2\hat{i}-\hat{j}+\hat{k}, \hat{i}+2\hat{j}-3\hat{k} \& 3\hat{i}+a\hat{j}+5\hat{k}$ are coplanar, then prove that $a=-4$.

Show that the vectors $\hat{i}-\hat{j}-2\hat{k}, 2\hat{i}+3\hat{j}+\hat{k} \& 7\hat{i}+3\hat{j}-4\hat{k}$ are coplanar.