A small particle moves to position $$5\hat{i} - 2\hat{j} + \hat{k}$$ from its initial position $$2\hat{i} + 3\hat{j} - 4\hat{k}$$ under the action of force $$5\hat{i} + 2\hat{j} + 7\hat{k}$$ N. The value of work done will be ______ J.

Let the initial position be $$\vec r_1 = 2\hat{i} + 3\hat{j} - 4\hat{k}$$ and the final position be $$\vec r_2 = 5\hat{i} - 2\hat{j} + \hat{k}$$.

Displacement is given by $$\vec d = \vec r_2 - \vec r_1$$. Substituting the values,

$$\vec d = (5 - 2)\hat{i} + (-2 - 3)\hat{j} + (1 - (-4))\hat{k} = 3\hat{i} - 5\hat{j} + 5\hat{k} \quad -(1)$$

The work done $$W$$ by a constant force $$\vec F$$ through displacement $$\vec d$$ is given by the formula

$$W = \vec F \cdot \vec d \quad -(2)$$

Given the force $$\vec F = 5\hat{i} + 2\hat{j} + 7\hat{k}$$, substitute into equation $$(2)$$:

$$W = (5\hat{i} + 2\hat{j} + 7\hat{k}) \cdot (3\hat{i} - 5\hat{j} + 5\hat{k})$$

Using the dot product rules $$\hat{i}\cdot\hat{i}=1$$, $$\hat{j}\cdot\hat{j}=1$$, $$\hat{k}\cdot\hat{k}=1$$ and zero for orthogonal unit vectors, we get

$$W = 5\times 3 + 2\times(-5) + 7\times 5 = 15 - 10 + 35 = 40\text{ J}$$.

Therefore, the work done is $$40\text{ J}$$.