$A$ vector $\vec{r}$ is inclined at equal angles to the three axes. If the magnitude of $\vec{r}$ is $2 \sqrt{3}$ units,find $\vec{r}$.

Given that the magnitude of the vector is $|\vec{r}| = 2 \sqrt{3}$.
Since $\vec{r}$ is equally inclined to the three axes,its direction cosines $l, m, n$ are equal,i.e.,$l = m = n$.
We know that the sum of the squares of direction cosines is $l^2 + m^2 + n^2 = 1$.
Substituting $l = m = n$,we get $l^2 + l^2 + l^2 = 1$,which implies $3l^2 = 1$.
Thus,$l^2 = \frac{1}{3}$,so $l = \pm \frac{1}{\sqrt{3}}$.
Since $l = m = n$,the unit vector $\hat{r}$ is given by $\hat{r} = \pm \frac{1}{\sqrt{3}} \hat{i} \pm \frac{1}{\sqrt{3}} \hat{j} \pm \frac{1}{\sqrt{3}} \hat{k}$.
Since $\vec{r} = |\vec{r}| \hat{r}$,we have $\vec{r} = 2 \sqrt{3} \left( \pm \frac{1}{\sqrt{3}} \hat{i} \pm \frac{1}{\sqrt{3}} \hat{j} \pm \frac{1}{\sqrt{3}} \hat{k} \right)$.
Therefore,$\vec{r} = \pm 2 \hat{i} \pm 2 \hat{j} \pm 2 \hat{k} = \pm 2(\hat{i} + \hat{j} + \hat{k})$.