$\begin{array}{l}\text{ If the vectors, }\overrightarrow{p}=(a+1)\hat{i}+\widehat{aj}+a\hat{k},\overrightarrow{q}=a\hat{i}+(a+1)\hat{j}+a\hat{k}\text{ and }\overrightarrow{r}=a\hat{i}+\hat{j}+(a+1)\hat{k} (a\in R)\text{ are }\\ \text{ coplanar and }3(\overrightarrow{p}·\overrightarrow{q}{)}^2-\lambda |\overrightarrow{r}\times \overline{q}{|}^2=0,\text{ then the value of }\lambda \text{ is }\end{array}$

$\text{ 3 vectors }\overline{p},\overline{q},\overline{r}\text{ are coplanar }\Rightarrow \left|\begin{array}{ccc}a+1& a& a\\ a& a+1& a\\ a& a& a+1\end{array}\right|=0$

Expand the determinant, 

$\begin{array}{c}\left(a+1\right)\left(a^2+2a+1-a^2\right)-a\left(a^2+a-a^2\right)+a\left(a^2-a^2-a\right)=0\\ \left(a+1\right)\left(2a+1\right)-a^2-a^2=0\\ 3a+1=0\\ a=-\frac{1}{3}\end{array}$

$\begin{array}{l}\text{ So, }\overline{p}=\frac{2}{3}\hat{i}-\frac{1}{3}\hat{j}-\frac{1}{3}\hat{k},\overline{q}=\frac{1}{3}(-\hat{i}+2\hat{j}-\hat{k}),\overline{r}=\frac{1}{3}(-\hat{i}-\hat{j}+2\hat{k})\\ \overline{p}·\overline{q}=\frac{1}{9}(-2-2+1)=-\frac{1}{3}\\ \overline{r}\times \overline{q}=\frac{1}{9}\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1& -1& 2\\ -1& 2& -1\end{array}\right|=\frac{1}{9}\left(\hat{i}\right(1-4)-\hat{j}(1+2)+\hat{k}(-2-1\left)\right)\\ =\frac{-1}{9}(3\hat{i}+3\hat{j}+3\hat{k})=\frac{-(\hat{i}+\hat{j}+\widehat{k)}}{3}\\ \lambda =\frac{3(\overline{p}·\overline{q}{)}^2}{|\overline{r}\times \overline{q}{|}^2}=1\end{array}$