Answer
$(80\sqrt 2, 80\sqrt 2-58)$
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 $ t=2$.
Therefore, $ x=(80 \cos 45^{\circ}) \times 2=80 \sqrt 2 \\ y=6+(80 \sin 45^{\circ})(2) -16(2)^2=80\sqrt 2-58$
So, the point is: $(80\sqrt 2, 80\sqrt 2-58)$
