Let  $\overrightarrow{a}=3\lambda \widehat{i}+(2\lambda +1)\widehat{j}+(\lambda +1)\widehat{k}$

$\overrightarrow{b}=(2\lambda -1)\widehat{i}+(2\lambda -1)\widehat{j}+(\lambda -2)\widehat{k}$

$\overrightarrow{c}=(4\lambda -1)\widehat{i}+3\lambda \widehat{j}+2\lambda \widehat{k}$
Product of all the values of  $\lambda$ is for which  $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are coplanar is 

a, b, c are coplanar 
$\Delta =\left|\begin{array}{ccc}3\lambda & 2\lambda +1& \lambda +1\\ 2\lambda -1& 2\lambda -1& \lambda -2\\ 4\lambda -1& 3\lambda & 2\lambda \end{array}\right|=0$
Use  $C_3\to C_3-C_2$  and  $C_2\to C_2-C_1$ to obtain
$\Delta =\left|\begin{array}{ccc}3\lambda & -\lambda +1& -\lambda \\ 2\lambda -1& 0& -\lambda -1\\ 4\lambda -1& -\lambda +1& -\lambda \end{array}\right|=0$
$R_1\to R_1-R_3$  gives 
$\Delta =\left|\begin{array}{ccc}-\lambda +1& 0& 0\\ 2\lambda -1& 0& -\lambda -1\\ 4\lambda -1& -\lambda +1& -\lambda \end{array}\right|=0$
$\begin{array}{l}\Delta =(-\lambda +1)(-\lambda +1)(\lambda +1)=0\\ \lambda =1,1,-1\end{array}$
Thus, product of values of  $\lambda$ is  $-1$.