The gravitational field in a region is given by $$\vec{g} = (5\hat{i} + 12\hat{j})$$ N kg$$^{-1}$$. The change in the gravitational potential energy of a particle of mass 2 kg when it is taken from the origin to a point (7 m, $$-3$$ m) is:

The gravitational field is given by $$\vec{g} = (5\hat{i} + 12\hat{j})$$ N kg$$^{-1}$$. We need to find the change in gravitational potential energy ($$\Delta U$$) for a particle of mass $$m = 2$$ kg when it is moved from the origin $$(0, 0)$$ to the point $$(7 \text{ m}, -3 \text{ m})$$.

The change in gravitational potential energy is given by $$\Delta U = -m \int_{\text{initial}}^{\text{final}} \vec{g} \cdot d\vec{r}$$. Since $$\vec{g}$$ is constant, the integral simplifies to a path-independent calculation.

Let $$d\vec{r} = dx\,\hat{i} + dy\,\hat{j}$$. Then, $$\vec{g} \cdot d\vec{r} = (5\hat{i} + 12\hat{j}) \cdot (dx\,\hat{i} + dy\,\hat{j}) = 5\,dx + 12\,dy$$.

So, the integral becomes:

$$$ \int_{(0,0)}^{(7,-3)} \vec{g} \cdot d\vec{r} = \int_{(0,0)}^{(7,-3)} (5\,dx + 12\,dy) $$$

We can compute this by integrating along the x-axis first and then the y-axis. From $$(0,0)$$ to $$(7,0)$$, $$y = 0$$ (so $$dy = 0$$), and then from $$(7,0)$$ to $$(7,-3)$$, $$x = 7$$ (so $$dx = 0$$).

First segment: from $$(0,0)$$ to $$(7,0)$$:

$$$ \int_{x=0}^{7} 5\,dx + \int_{y=0}^{0} 12\,dy = \int_{0}^{7} 5\,dx + 0 = 5 \left[ x \right]_{0}^{7} = 5 \times (7 - 0) = 35 $$$

Second segment: from $$(7,0)$$ to $$(7,-3)$$:

$$$ \int_{x=7}^{7} 5\,dx + \int_{y=0}^{-3} 12\,dy = 0 + \int_{0}^{-3} 12\,dy = 12 \left[ y \right]_{0}^{-3} = 12 \times (-3 - 0) = 12 \times (-3) = -36 $$$

Adding both segments:

$$$ \int_{(0,0)}^{(7,-3)} \vec{g} \cdot d\vec{r} = 35 + (-36) = -1 \text{ J kg}^{-1} $$$

Alternatively, using a straight-line path parameterization: let $$x = 7t$$, $$y = -3t$$, where $$t$$ goes from 0 to 1. Then $$dx = 7\,dt$$, $$dy = -3\,dt$$.

$$$ \vec{g} \cdot d\vec{r} = 5(7\,dt) + 12(-3\,dt) = 35\,dt - 36\,dt = -dt $$$

$$$ \int_{0}^{1} (-1)\,dt = -1 \left[ t \right]_{0}^{1} = -1 \times (1 - 0) = -1 \text{ J kg}^{-1} $$$

Both methods give the same result: $$-1$$ J kg$$^{-1}$$.

Now, the change in potential energy is:

$$$ \Delta U = -m \times \left( \int \vec{g} \cdot d\vec{r} \right) = - (2 \text{ kg}) \times (-1 \text{ J kg}^{-1}) = -2 \times (-1) = 2 \text{ J} $$$

Hence, the change in gravitational potential energy is 2 J.

Comparing with the options:

A. 71 J

B. $$13\sqrt{58}$$ J

C. 2 J

D. 1 J

So, the correct answer is Option C.