Find the value of λ for which the four points with position vectors -j ̂-k ̂,4i ̂+5j ̂+λk ̂,3i ̂+9j ̂+4k ̂ and -4i ̂+4j ̂+4k ̂ are coplanar.

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ \overrightarrow{\mathrm{OA}}=-\hat{\mathrm{j}}-\hat{\mathrm{k}} \\ \overrightarrow{\mathrm{OB}}=4\hat{\mathrm{i}}+5\hat{\mathrm{j}}+\mathrm{\lambda }\hat{\mathrm{k}} \\ \overrightarrow{\mathrm{OC}}=3\hat{\mathrm{i}}+9\hat{\mathrm{j}}+4\hat{\mathrm{k}} \\ \overrightarrow{\mathrm{OD}}=-4\hat{\mathrm{i}}+4\hat{\mathrm{j}}+4\hat{\mathrm{k}} \\ \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=4\hat{\mathrm{i}}+5\hat{\mathrm{j}}+\mathrm{\lambda }\hat{\mathrm{k}}-\left(-\hat{\mathrm{j}}-\hat{\mathrm{k}}\right)=4\hat{\mathrm{i}}+6\hat{\mathrm{j}}+\left(\mathrm{\lambda }+1\right)\hat{\mathrm{k}} \\ \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}=3\hat{\mathrm{i}}+9\hat{\mathrm{j}}+4\hat{\mathrm{k}}-\left(-\hat{\mathrm{j}}-\hat{\mathrm{k}}\right)=3\hat{\mathrm{i}}+10\hat{\mathrm{j}}+5\hat{\mathrm{k}} \\ \overrightarrow{\mathrm{AD}}=\overrightarrow{\mathrm{OD}}-\overrightarrow{\mathrm{OA}}=\left(-4\hat{\mathrm{i}}+4\hat{\mathrm{j}}+4\hat{\mathrm{k}}\right)-\left(-\hat{\mathrm{j}}-\hat{\mathrm{k}}\right)=-4\hat{\mathrm{i}}+5\hat{\mathrm{j}}+5\hat{\mathrm{k}} \\ \mathrm{As} \overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AC}} \mathrm{and} \overrightarrow{\mathrm{AD}} \mathrm{are} \mathrm{coplanar}, \mathrm{hence} \mathrm{their} \mathrm{scalar} \mathrm{triple} \mathrm{product} \mathrm{will} \mathrm{be} 0. \\ \Rightarrow \overrightarrow{\mathrm{AB}}.\left(\overrightarrow{\mathrm{AC}}\times \overrightarrow{\mathrm{AD}}\right)=0 \\ \Rightarrow \left|\begin{array}{ccc}4& 6& \left(\mathrm{\lambda }+1\right)\\ 3& 10& 5\\ -4& 5& 5\end{array}\right|=0 \\ \mathrm{Expanding} \mathrm{along} {\mathrm{R}}_1, \mathrm{we} \mathrm{get}: \\ 4\left(50-25\right)-6\left(15+20\right)+\left(\mathrm{\lambda }+1\right)\left(15+40\right)=0 \\ 100-210+\left(\mathrm{\lambda }+1\right)\left(55\right)=0 \\ \left(\mathrm{\lambda }+1\right)\left(55\right)=110 \\ \Rightarrow \mathrm{\lambda }+1=\frac{110}{55}=2 \\ \Rightarrow \mathbf{\lambda }\mathbf{=}\mathbf{1}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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