If $\left(1,5,35\right),\left(7,5,5\right),\left(1,\lambda ,7\right)$ and $\left(2\lambda ,1,2\right)$ are coplanar, then the sum of all possible values of $\lambda$ is:

$\overline{AB}=6\hat{i}-30\hat{k}, \overline{BC}=-6\widehat{i+}\left(\lambda -5\right)\hat{j}+2\hat{k}$

$\overrightarrow{CD}=\left(2\lambda -1\right)\hat{i}+\left(1-\lambda \right)\hat{j}-5\hat{k}$

Points are coplanar
$\Rightarrow 0=\left|\begin{array}{ccc}6& 0& -30\\ -6& \lambda -5& 2\\ 2\lambda -1& 1-\lambda & -5\end{array}\right|$

$=6\left(-5\lambda +25-2+2\lambda \right)$

$-30\left(-6+6\lambda -\left(2{\lambda }^2-\lambda -10\lambda +5\right)\right)$

$=6\left(-3\lambda +23\right)-30\left(-2{\lambda }^2+11\lambda -5-6+6\lambda \right)$

$=6\left(-3\lambda +23\right)-30\left(-2{\lambda }^2+17\lambda -11\right)$

$=6\left(-3\lambda +23+10{\lambda }^2-85\lambda +55\right)$

$=6\left(10{\lambda }^2-88\lambda +78\right)=12\left(5{\lambda }^2-44\lambda +39\right)$

$\Rightarrow 0=12\left(5{\lambda }^2-44\lambda +39\right)$

${\lambda }_1+{\lambda }_2=\frac{44}{5}$