Then the tangent plane l equation of the curved surface F(x, y, z) is: f ′ x (x0, y0, z0) (x-x0)+f ′ y (x0, y0, z0) (y-y0)+f ′ z (x0, y0, z0) (z).

Where P(x0, y0, z0) is the tangent point, and f ′ x, f ′ y and f ′ z are the partial derivatives of F(x, y, z) to x, y and z respectively.

You can get f ′ x = 2x0f ′ y = 2y0f ′ z =1.

The tangent plane L is parallel to the plane T, so 2x0/2 = 2Y0/2 =11.

∴ x0 = 1 y0 =1z0 = 4-(x0) 2-(y0) 2 = 2, that is, P( 1,1,2).

So the section L: 2 (x-1)+2 (y-1)+(z-2) = 0 means 2x+2y+z-6=0.

2) Take the point Q (0,0,0) on the plane T.

Vector PQ=( 1, 1, 2)

The normal vector of plane T is: vector n = {2,2, 1}

Then the shortest distance s from the surface to the plane t is the module of the projection of the vector PQ to the normal vector n.

∴S= vector pq cos < pq, n >

= (Vector PQ，Vector n)/(│ Vector n│)

=( 1×2+ 1×2+2× 1)/√(2^2+2^2+ 1^2)

=2

Where cos < PQ, n > represents the cosine of the included angle (acute angle) between vector PQ and vector n.

(Note: the longest distance is ∞, which should be the shortest distance)