Let for a triangle ABC → AB = - 2 ^ i + ^ j + 3 ^ k → CB = α ^ i +...

Given,

A triangle $ABC$

\overset{\to}{AB} = - 2\overset{^}{i} + \overset{^}{j} + 3\overset{^}{k}

\overset{\to}{CB} = \alpha\overset{^}{i} + \beta\overset{^}{j} + \gamma\overset{^}{k}

\overset{\to}{CA} = 4\overset{^}{i} + 3\overset{^}{j} + \delta\overset{^}{k}

Now plotting the diagram we get,

Now from triangle law of addition we get,

$\overset{\to}{CA} + \overset{\to}{AB} = \overset{\to}{CB}$

⇒ 4\overset{^}{i} + 3\overset{^}{j} + \delta\overset{^}{k} + \left(- 2\overset{^}{i} + \overset{^}{j} + 3\overset{^}{k}\right) = \alpha\overset{^}{i} + \beta\overset{^}{j} + \gamma\overset{^}{k}

⇒ 2\overset{^}{i} + 4\overset{^}{j} + \left(\delta + 3\right)\overset{^}{k} = \alpha\overset{^}{i} + \beta\overset{^}{j} + \gamma\overset{^}{k}

Now on comparing both side we get,

$\alpha = 2, \beta = 4 \text{ and } \gamma = \delta + 3$

Now area of triangle is given by,

$A = \dfrac{1}{2}\left|\overset{\to}{AB} \times \overset{\to}{BC}\right|$

⇒ 5\sqrt{6} = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ - 2 & 1 & 3 \\ 2 & 4 & \gamma\end{cases}\right|

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

$⇒ 5{\gamma}^{2} = 320$

$⇒ {\gamma}^{2} = 64 ⇒ \gamma = 8$

So, $\delta = 8 - 3 = 5$

\overset{\to}{CB} \cdot \overset{\to}{CA} = \left(\alpha\overset{^}{i} + \beta\overset{^}{j} + \gamma\overset{^}{k}\right) \cdot \left(4\overset{^}{i} + 3\overset{^}{j} + \delta\overset{^}{k}\right)

⇒ \overset{\to}{CB} \cdot \overset{\to}{CA} = \left(2\overset{^}{i} + 4\overset{^}{j} + 8\overset{^}{k}\right) \cdot \left(4\overset{^}{i} + 3\overset{^}{j} + 5\overset{^}{k}\right) = 8 + 12 + 40 = 60