If the volume of the tetrahedron whose vertices are $(1,-6,10),(1,-3,7)(5,-1,\lambda )\text{ and }(7,-4,7)$, is $11$ cubic units, then the sum of the values of $\lambda$, is 

The vertices are $A(1,-6,10),B(-1,-3,7)$,$C(5,-1,\lambda )\text{ and }D(7,-4,7)$

$\begin{array}{l}\therefore \overrightarrow{AB}=-2\widehat{i}+3\widehat{j}-3\widehat{k},\overrightarrow{AC}=4\widehat{i}+5\widehat{j}+(\lambda -10)\widehat{k}\\ \overrightarrow{AD}=6\widehat{i}+2\widehat{j}-3\widehat{k}\end{array}$

It is given that the volume of the tetrahedron $ABCD$ is $11$ cubic units. 

$\begin{array}{l}\frac{1}{6}\left[\begin{array}{ccc}\overrightarrow{AB}& \overrightarrow{AC}& \overrightarrow{AD}\end{array}\right]=\pm 11\\ \Rightarrow \left|\begin{array}{ccc}-2& 3& -3\\ 4& 5& \lambda -10\\ 6& 2& -3\end{array}\right|=\pm 66\end{array}$

$\Rightarrow \left|\begin{array}{ccc}-2& 3& -3\\ 0& 0& \lambda -4\\ 6& 2& -3\end{array}\right|=\pm 66$       Applying $R_2\to R_2-\left(R_1+R_3\right)$

$\begin{array}{l}\Rightarrow -(\lambda -4)(-4-18)=\pm 66\\ \Rightarrow 22(\lambda -4)=\pm 66\\ \Rightarrow \lambda -4=\pm 3\Rightarrow \lambda =7,1\end{array}$

Hence, sum of the values of $\lambda$ is $7+1=8$