Answer
$(60 \sqrt 3,1)$
Work Step by Step
The line passing through the point $ P(a,b,c)$ and parallel to the normal vector $ n=\lt p,q,r\gt $ can be represented by the parametric equations as shown below:
$ x=a+pt; y=b+qt; z=c+qt $
Here, $ t \to $ Real number.
Now, plug in the data.
Therefore, $ x=120 \cos 30^{\circ}=120 \times \dfrac{\sqrt 3}{2} =60 \sqrt 3 \\ y=120 \sin 30^{\circ}-59=1$
So, the point is: $(60 \sqrt 3,1)$
