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Since the line makes equal angles with the coordinate axes, the direction cosines are equal in magnitude. Let the direction cosines be $l, m, n$. Then $l = m = n$ or $l = m = -n$ (and other sign combinations). Since $l^2 + m^2 + n^2 = 1$, we have $l^2 + l^2 + l^2 = 1$, which gives $3l^2 = 1$, so $l = \pm \frac{1}{\sqrt{3}}$. Thus, the direction cosines are $\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$ or $\left(-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)$ or $\left(\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)$ and so on.
The direction ratios are proportional to the direction cosines. We can take the direction ratios as $(1, 1, 1)$ or $(-1, -1, -1)$ or $(1, -1, -1)$ and so on. We can choose any of these sets of direction ratios.
The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is given by $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\lambda$ is a scalar. The position vector of the point (2, 3, -5) is $\vec{a} = 2\hat{i} + 3\hat{j} - 5\hat{k}$. We can take the direction vector $\vec{b}$ as $\hat{i} + \hat{j} + \hat{k}$ or $-\hat{i} - \hat{j} - \hat{k}$ or $\hat{i} - \hat{j} - \hat{k}$ and so on. Therefore, the vector equation of the line is $\vec{r} = (2\hat{i} + 3\hat{j} - 5\hat{k}) + \lambda (\hat{i} + \hat{j} + \hat{k})$ or $\vec{r} = (2\hat{i} + 3\hat{j} - 5\hat{k}) + \lambda (-\hat{i} - \hat{j} - \hat{k})$ or $\vec{r} = (2\hat{i} + 3\hat{j} - 5\hat{k}) + \lambda (\hat{i} - \hat{j} - \hat{k})$ and so on.
Final Answer: $\vec{r} = (2\hat{i} + 3\hat{j} - 5\hat{k}) + \lambda (\hat{i} + \hat{j} + \hat{k})$ (or equivalent)
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