Find the vector equation of the line passing through the point $(2, 3, -5)$ and making equal angles with the coordinate axes.

$\vec{r} = (2\hat{i} + 3\hat{j} - 5\hat{k}) \pm \frac{k}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})$

The correct answer is Option (2) → $\vec{r} = (2\hat{i} + 3\hat{j} - 5\hat{k}) \pm \frac{k}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})$ ##

Given $\alpha = \beta = \gamma$

Now, $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$

$3\cos^2\alpha = 1 \text{ or, } 3\cos^2\beta = 1$

$3\cos^2\alpha = 1 \text{ or, } \cos^2\alpha = \frac{1}{3} \quad [∵\alpha = \beta = \gamma]$

or, $\cos\alpha = \pm \frac{1}{\sqrt{3}}$

Thus, $\cos\alpha = \cos\beta = \cos\gamma = \pm \frac{1}{\sqrt{3}}$

$= l = m = n$

Let the required equation of line is

$\vec{r} = \vec{a} + k\vec{b}$

Here,

$\vec{a} = 2\hat{i} + 3\hat{j} - 5\hat{k}$

$\vec{b} = \pm \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})$

Thus, the required equation is

$\vec{r} = (2\hat{i} + 3\hat{j} - 5\hat{k}) \pm \frac{k}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})$