Vector se = (0,2,4), vector SF=( 1, 4,4),

The vector SE*n=2y-4z=0

The vector SF*n=x+4y-4z=0.

Let x= 1, and solve the above two equations simultaneously to get y =- 1/2 and z =- 1/4.

∴ Normal vector n = (1,-1/2,-1/4)

A plane has countless normal vectors, just pick one.

2. Distance from point to plane

Let the normal vector a=(x 1, y 1, z 1) of a given plane ABC, d be a point out of the plane, and the vector AD=(x2, y2, z2).

Then the distance from d to plane ABC is the projection of vector AD on the plane normal.

That's because

∴d=| vector a* vector AD|/| vector a|

∫ vector as = (0 0,0,4), | vector n | = √ (1+1/4+1/6) = √ 21/4.

∴ distance from point A to plane SEF d=| vector n* vector AS|/| vector n|

=|( 1,- 1/2,- 1/4)*(0,0,4)|/( √2 1/4)=( 1/ 16)/( √2 1/4)= 4√2 1/2 1

Answer supplementary questions:

Establish a space rectangular coordinate system A-xyz with A AS the origin, AD as the X axis, AB as the steering wheel as the Y axis and AS as the positive direction of the Z axis.

Then point coordinates:

A(0，0，0)，E(0， 1/2，0)，F( 1，4，0)，S(0，0，4)

The vector SE=(0, 1/2, -4), the vector SF=( 1, 4, -4) and the vector SA=(0, 0, -4).

Let the vector n=(x, y, z) be the normal vector of the plane SEF.

1/2y-4z=0

X+4y-4z=0

Let x =1= = > 7/2y+ 1=0== >y=-2/7，z= 1/28

∴ Vector n = (1, -2/7, 1/28)= > | Vector n|=√849/28

Vector SA* Vector n=0+0- 1/7=- 1/7.

D=| Vector SA* Vector n |/| Vector n|=4√849/849