Well, that's pretty straight forward isn't it? Rewrite this as \(\displaystyle f(x,y,z)= x^2+ z^2- y= 0\) so you can think of this paraboloid as being a "level surface" for f(x,y,z). What is the gradient of f at any point (x,y,z)? What is a normal vector to the plane? where are those two vectors parallel?

Gradient vector of f is \(\displaystyle <2x,-1,2z>\)

Normal to the plane is \(\displaystyle <1,2,7>\)

Tangent plane is parallel to that plane if the corresponding normal vectors are parallel, so

\(\displaystyle <2x_o,-1,2z_o> = c<1,2,7>\)

or equivalently \(\displaystyle <x_o,\frac{-1}{2},z_o> = k<1,2,7>\)

\(\displaystyle x_o = k \)

\(\displaystyle \frac{-1}{2} = 2k \)

\(\displaystyle z_o = 7k \)

Solving for k yields \(\displaystyle k=\frac{-1}{4}, x_o=\frac{-1}{4}, z_o=\frac{-7}{4}, y_o=\frac{50}{16} \)

Did I do that correctly?