Answer:
A point on the ellipsoid is (-4,2,2) or (4,-2,-2)
Step-by-step explanation:
Given equation of ellipsoid f(x,y,z) :
Parametric equations:
x=-4t-1
y=2t+1
z=8t+3
Finding the gradient of function

So, The directions vectors=(-4,2,8)
Now the line is perpendicular to plane when direction vector is parallel to the normal vector of line

So, 


Substitute the value of x , y and z in the ellipsoid equation

With 
x=-2(2)=-4
y=2
z=2
With
x=-2(-2)=4
y=-2
z=-2
Hence a point on the ellipsoid is (-4,2,2) or (4,-2,-2)