Let the vector $\mathbf{v}$ make an angle $\alpha$ with each of the three axes, then direction cosine of $\mathbf{v}$ are $ < \cos \alpha, \cos \alpha, \cos \alpha>$
Also, $\quad \cos ^2 \alpha+\cos ^2 \alpha+\cos ^2 \alpha=1$
Hence, direction cosine of $\mathbf{v}$ are $ < \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}>$ or $ < -\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}>$
So, the required line which makes equal angle with the coordinate axes is
$\mathbf{v}= \pm \frac{1}{\sqrt{3}} \mathbf{i} \pm \frac{1}{\sqrt{3}} \mathbf{j} \pm \frac{1}{\sqrt{3}} \mathbf{k}$
Now, the magnitude of the projection of the vector $\mathbf{a}=4 \mathbf{i}-3 \mathbf{j}+2 \mathbf{j}$ on line $\mathbf{v}$.
$\therefore$ Projection of a along r
$\begin{aligned}
\mathbf{v} & =\frac{\mathbf{a} \cdot \mathbf{v}}{|\mathbf{v}|} \\
& =\frac{3 / \sqrt{3}}{1}=\sqrt{3}
\end{aligned}$