Given that

$\overrightarrow{a}=\lambda \hat{i}+\hat{j}-\hat{k}$

$\overrightarrow{b}=3\hat{i}-\hat{j}+2\hat{k}$

Also $\left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right)\times \overrightarrow{c}=\overrightarrow{0}$

$\Rightarrow k\left(\overrightarrow{a}+\overrightarrow{b}\right)=\overrightarrow{c}$

$\overrightarrow{a}·\overrightarrow{c}=-17$ and $\overrightarrow{b}·\overrightarrow{c}=-20$

Now, Take $\overrightarrow{a}·\overrightarrow{c}=-17$

$\Rightarrow k\left(\lambda \hat{i}+\hat{j}-\hat{k}\right)·\left(\lambda \hat{i}+\hat{j}-\hat{k}+3\hat{i}-\hat{j}+2\hat{k}\right)=-17$

$\Rightarrow k\left({\lambda }^2+3\lambda +0-1\right)=-17$

 $\Rightarrow k\left({\lambda }^2+3\lambda -1\right)=-17 ....\left(1\right)$

Similarly, on taking $\overrightarrow{b}·\overrightarrow{c}=-20$, we get, 

$k\left(3\hat{i}-\hat{j}+2\hat{k}\right)·\left(\lambda \hat{i}+\hat{j}-\hat{k}+3\hat{i}-\hat{j}+2\hat{k}\right)=-20$

$\Rightarrow k\left(3\lambda +9+2\right)=-20$

$\Rightarrow k\left(3\lambda +11\right)=-20 ...\left(2\right)$

Now on solving equation $\left(1\right) \& \left(2\right)$ we get,

$\Rightarrow 20{\lambda }^2+9\lambda -207=0$

$\Rightarrow \lambda =3, \frac{-69}{20}$

For $\lambda =3, k=-1$

$\Rightarrow \overrightarrow{c}=-1\left(\overrightarrow{a}+\overrightarrow{b}\right)$

$\Rightarrow -\left(\left(\lambda +3\right)\hat{i}+\hat{k}\right)=-6\hat{i}-\hat{k}$

$\Rightarrow \overrightarrow{c}\times \left(\lambda \hat{i}+\hat{j}+\hat{k}\right)=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -6& 0& -1\\ 3& 1& 1\end{array}\right|=\hat{i}-3\hat{j}+6\hat{k}$

$\Rightarrow {\left|\overrightarrow{c}\times \left(\lambda \hat{i}+\hat{j}+\hat{k}\right)\right|}^2=46$

Hence this is the correct option.