Given, position vectors of the points $A, B, C$ and $D$ be

$5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \lambda \hat{\mathbf{k}}, \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}},-2 \hat{\mathbf{i}}+\lambda \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ and $-\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$

$$ \begin{aligned} \overrightarrow{A B} & =(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})-(5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \lambda \hat{\mathbf{k}})=-4 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+(3-2 \lambda) \hat{\mathbf{k}} \\\\ \overrightarrow{A C} & =(-2 \hat{\mathbf{i}}+\lambda \hat{\mathbf{j}}+4 \hat{\mathbf{k}})-(5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \lambda \hat{\mathbf{k}}) \\\\ & =-7 \hat{\mathbf{i}}+(\lambda-5) \hat{\mathbf{j}}+(4-2 \lambda) \hat{\mathbf{k}} \end{aligned} $$

$$ \begin{aligned} \text { and } \overrightarrow{A D} & =(-\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})-(5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \lambda \hat{\mathbf{k}}) \\\\ & =-6 \hat{\mathbf{i}}+(6-2 \lambda) \hat{\mathbf{k}} \end{aligned} $$

Since, points $A, B, C$ and $D$ are coplanar

$$ \begin{aligned} & \therefore [ { \overrightarrow{AB}~ \overrightarrow{AC}~ \overrightarrow{AD}] }=0 \\\\ & \Rightarrow \left|\begin{array}{ccc} -4 & -3 & (3-2 \lambda) \\ -7 & (\lambda-5) & (4-2 \lambda) \\ -6 & 0 & 6-2 \lambda \end{array}\right|=0 \end{aligned} $$

$$ \begin{aligned} & \Rightarrow -6\left\{(-12+6 \lambda)-\left(13 \lambda-15-2 \lambda^2\right)\right\} \\\\ & +(6-2 \lambda)\{-4 \lambda+20-21\} =0 \\\\ & \Rightarrow -6\left(2 \lambda^2-7 \lambda+3\right)+(6-2 \lambda)(-4 \lambda-1) =0 \\\\ & \Rightarrow -12 \lambda^2+42 \lambda-18+8 \lambda^2-22 \lambda-6 =0 \\\\ & \Rightarrow -4 \lambda^2+20 \lambda-24 =0 \\\\ & \Rightarrow \lambda^2-5 \lambda+6 =0 \\\\ & \Rightarrow (\lambda-2)(\lambda-3) =0 \\\\ & \Rightarrow \lambda =2,3 \end{aligned} $$

$$ \therefore \sum\limits_{\lambda \varepsilon S}(\lambda+2)^2=(2+2)^2+(3+2)^2=16+25=41 $$

Vector Algebra

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