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Let the vector $\vec{r}$ be inclined at an angle $\alpha$ with each of the x, y, and z axes. Then, the direction cosines of $\vec{r}$ are $l = \cos\alpha$, $m = \cos\alpha$, and $n = \cos\alpha$.
We know that $l^2 + m^2 + n^2 = 1$. Substituting the values of $l, m, n$, we get: $$(\cos\alpha)^2 + (\cos\alpha)^2 + (\cos\alpha)^2 = 1$$ $$3\cos^2\alpha = 1$$ $$\cos^2\alpha = \frac{1}{3}$$ $$\cos\alpha = \pm\frac{1}{\sqrt{3}}$$
Thus, the direction cosines are $l = \pm\frac{1}{\sqrt{3}}$, $m = \pm\frac{1}{\sqrt{3}}$, and $n = \pm\frac{1}{\sqrt{3}}$.
Given that the magnitude of $\vec{r}$ is $|\vec{r}| = 5\sqrt{3}$. The vector $\vec{r}$ can be written as: $$\vec{r} = |\vec{r}|(l\hat{i} + m\hat{j} + n\hat{k})$$ $$\vec{r} = 5\sqrt{3}\left(\pm\frac{1}{\sqrt{3}}\hat{i} \pm \frac{1}{\sqrt{3}}\hat{j} \pm \frac{1}{\sqrt{3}}\hat{k}\right)$$ $$\vec{r} = \pm 5\hat{i} \pm 5\hat{j} \pm 5\hat{k}$$
Therefore, $\vec{r} = \pm 5(\hat{i} + \hat{j} + \hat{k})$.
Final Answer: $\vec{r} = \pm 5(\hat{i} + \hat{j} + \hat{k})$
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