If the volume of tetrahedron, whose vertices are with position vectors \(\hat{i}-6 \hat{j}+10 \hat{k},-\hat{i}-3 \hat{j}+7 \hat{k}\), \(5 \hat{i}-\hat{j}+\lambda \hat{k}\) and \(7 \hat{i}-4 \hat{j}+7 \hat{k}\) is 11 cubic units, then value of \(\lambda\) is

\(\begin{aligned} & \text { Let } \overline{\mathrm{a}}=\hat{\mathrm{i}}-6 \hat{\mathrm{j}}+10 \hat{\mathrm{k}} \text {, } \\ & \overline{\mathrm{b}}=-\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+7 \hat{\mathrm{k}} \text {, } \\ & \overline{\mathrm{c}}=5 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\lambda \hat{\mathrm{k}} \text {, } \\ & \overline{\mathrm{d}}=7 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+7 \hat{\mathrm{k}} \\ & \therefore \quad \overline{\mathrm{AB}}=-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \\ & \overline{\mathrm{AC}}=4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+(\lambda-10) \hat{\mathrm{k}} \\ & \overline{\mathrm{AD}}=6 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \\ & \therefore \quad \text { Volume of tetrahedron }=\frac{1}{6}\left[\begin{array}{lll}\overline{\mathrm{AB}} & \overline{\mathrm{AC}} & \overline{\mathrm{AD}}\end{array}\right] \\ & \Rightarrow 11=\frac{1}{6}\left|\begin{array}{ccc}-2 & 3 & -3 \\ 4 & 5 & \lambda-10 \\ 6 & 2 & -3\end{array}\right| \\ & \Rightarrow 11=\frac{1}{6}\{-2(-15-2 \lambda+20)-3(-12-6 \lambda+60) \\ & -3(8-30)\} \\ & \Rightarrow \lambda=7& \end{aligned}\)