Let for a triangle A B C A B → = - 2 i ^ + j ^ + 3 k ^ C B → = α i ^ + β j ^ + γ…

Given,

A triangle $ABC$ 

$\overrightarrow{AB}=-2\hat{i}+\hat{j}+3\hat{k}$

$\overrightarrow{CB}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$

$\overrightarrow{CA}=4\hat{i}+3\hat{j}+\delta \hat{k}$

Now plotting the diagram we get,

Now from triangle law of addition we get,

$\overrightarrow{CA}+\overrightarrow{AB}=\overrightarrow{CB}$

$\Rightarrow 4\hat{i}+3\hat{j}+\delta \hat{k}+\left(-2\hat{i}+\hat{j}+3\hat{k}\right)=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$

$\Rightarrow 2\hat{i}+4\hat{j}+\left(\delta +3\right)\hat{k}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$

Now on comparing both side we get,

$\alpha =2, \beta =4 \& \gamma =\delta +3$

Now area of triangle is given by,

$A=\frac{1}{2}\left|\overrightarrow{AB}\times \overrightarrow{BC}\right|$

$\Rightarrow 5\sqrt{6}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -2& 1& 3\\ 2& 4& \gamma \end{array}\right|$

$\Rightarrow {\left(5\sqrt{6}\right)}^2={\left(\gamma -12\right)}^2+{\left(6+2\gamma \right)}^2+100$

$\Rightarrow 5{\gamma }^2=320$

$\Rightarrow {\gamma }^2=64\Rightarrow \gamma =8$

So, $\delta =8-3=5$

$\overrightarrow{CB}·\overrightarrow{CA}=\left(\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}\right)·\left(4\hat{i}+3\hat{j}+\delta \hat{k}\right)$

$\Rightarrow \overrightarrow{CB}·\overrightarrow{CA}=\left(2\hat{i}+4\hat{j}+8\hat{k}\right)·\left(4\hat{i}+3\hat{j}+5\hat{k}\right)=8+12+40=60$