If the volume of tetrahedron whose vertices are \(\mathrm{A} \equiv(1,-6,10), \mathrm{B} \equiv(-1,-3,7), \mathrm{C} \equiv(5,-1, \mathrm{k})\) and \(D \equiv(7,-4,7)\) is 11 cu . units, then the value of \(k\) is

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