The force exerted by an electric charge at the origin on a charged particle at the point $(x, y, z)$ with position vector $\vec{r}=\langle x, y, z)$ is $\vec{F}(\vec{r})=\frac{K \vec{r}}{|\vec{r}|^{3}}$, where $K$ is constant. Assume $K=25$. Find the work done as the particle moves along a straight line from $(2,0,0)$ to $(2,3,5)$.

Step 1 ：Given the force exerted by an electric charge at the origin on a charged particle at the point \((x, y, z)\) with position vector \(\vec{r}=\langle x, y, z \rangle\) is \(\vec{F}(\vec{r})=\frac{K \vec{r}}{|\vec{r}|^{3}}\), where \(K\) is a constant and \(K=25\).

Step 2 ：We need to find the work done as the particle moves along a straight line from \((2,0,0)\) to \((2,3,5)\).

Step 3 ：The work done by a force field \(\vec{F}\) along a curve \(C\) from point \(A\) to point \(B\) is given by the line integral \(\int_C \vec{F} \cdot d\vec{r}\).

Step 4 ：Here, \(\vec{F}(\vec{r})=\frac{K \vec{r}}{|\vec{r}|^{3}}\) is the force field, and the curve \(C\) is a straight line from \((2,0,0)\) to \((2,3,5)\).

Step 5 ：We can parameterize this line as \(\vec{r}(t) = \langle 2, 3t, 5t \rangle\) for \(0 \leq t \leq 1\). Then, \(d\vec{r} = \langle 0, 3, 5 \rangle dt\).

Step 6 ：We can substitute these into the line integral and compute it to find the work done.

Step 7 ：Let's calculate the integral: \(\int_0^1 \frac{K \vec{r}(t)}{|\vec{r}(t)|^{3}} \cdot d\vec{r}\).

Step 8 ：Substituting the values, we get \(\int_0^1 \frac{25 \langle 2, 3t, 5t \rangle}{|\langle 2, 3t, 5t \rangle|^{3}} \cdot \langle 0, 3, 5 \rangle dt\).

Step 9 ：Solving the integral, we get \(\frac{450\sqrt{38}}{323} + \frac{400}{17}\).

Step 10 ：Thus, the work done as the particle moves along a straight line from \((2,0,0)\) to \((2,3,5)\) is \(\boxed{\frac{450\sqrt{38}}{323} + \frac{400}{17}}\).