Let for a triangle $$ABC$$
$$\vec{AB} = -2\hat{i} + \hat{j} + 3\hat{k}$$
$$\vec{CB} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$$
$$\vec{CA} = 4\hat{i} + 3\hat{j} + \delta\hat{k}$$
If $$\delta > 0$$ and the area of the triangle $$ABC$$ is $$5\sqrt{6}$$ then $$\vec{CB} \cdot \vec{CA}$$ is equal to

Given: $$\vec{AB} = -2\hat{i} + \hat{j} + 3\hat{k}$$, $$\vec{CA} = 4\hat{i} + 3\hat{j} + \delta\hat{k}$$

$$\vec{CB} = \vec{CA} + \vec{AB} = (4-2)\hat{i} + (3+1)\hat{j} + (\delta+3)\hat{k} = 2\hat{i} + 4\hat{j} + (\delta+3)\hat{k}$$

Area of triangle = $$\frac{1}{2}|\vec{CA} \times \vec{CB}| = 5\sqrt{6}$$

$$\vec{CA} \times \vec{CB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & \delta \\ 2 & 4 & \delta+3 \end{vmatrix}$$

$$= \hat{i}(3(\delta+3) - 4\delta) - \hat{j}(4(\delta+3) - 2\delta) + \hat{k}(16-6)$$

$$= \hat{i}(3\delta+9-4\delta) - \hat{j}(4\delta+12-2\delta) + 10\hat{k}$$

$$= (9-\delta)\hat{i} - (2\delta+12)\hat{j} + 10\hat{k}$$

$$|\vec{CA} \times \vec{CB}|^2 = (9-\delta)^2 + (2\delta+12)^2 + 100$$

Area² = $$\frac{1}{4}[(9-\delta)^2 + (2\delta+12)^2 + 100] = 150$$

$$(9-\delta)^2 + (2\delta+12)^2 + 100 = 600$$

$$81-18\delta+\delta^2 + 4\delta^2+48\delta+144 + 100 = 600$$

$$5\delta^2 + 30\delta + 325 = 600$$

$$5\delta^2 + 30\delta - 275 = 0$$

$$\delta^2 + 6\delta - 55 = 0$$

$$(\delta+11)(\delta-5) = 0$$

Since $$\delta > 0$$: $$\delta = 5$$

$$\vec{CB} = 2\hat{i} + 4\hat{j} + 8\hat{k}$$, $$\vec{CA} = 4\hat{i} + 3\hat{j} + 5\hat{k}$$

$$\vec{CB} \cdot \vec{CA} = 8 + 12 + 40 = 60$$

This matches option 1: 60.