The distance of origin from line $\left(1+\sqrt{3}\right)\mathrm{y}+\left(1-\sqrt{3}\right)\mathrm{x}=10$ measured along the line $\mathrm{y}=\sqrt{3}\mathrm{x}+4$ is 

Slope of the line $\mathrm{L}\equiv \left(1+\sqrt{3}\right)\mathrm{y}+\left(1-\sqrt{3}\right)\mathrm{x}=10$ is 

$\frac{\sqrt{3}-1}{\sqrt{3}+1}=\tan {15}^0$

Slope of line $\mathrm{y}=\sqrt{3}\mathrm{x}+\mathrm{k}$ is $\sqrt{3}=\tan {60}^0$

In the figure, distance of origin from line L measured along the line $\mathrm{y}=\sqrt{3}\mathrm{x}+\mathrm{k}$ is OP

From the figure,

$\mathrm{OP}=\mathrm{OM}\sec {45}^0$

$=\frac{\left|0+0-10\right|}{{\left(\sqrt{3}-1\right)}^2+{\left(\sqrt{3}+1\right)}^2}\sqrt{2}=\frac{10}{\sqrt{8}}\sqrt{2}=5$