Vector AC = [2,-1, 0]

Cross product of vector AB and vector AC

= [2*0 - (- 1)*(- 1), 2*(- 1) - 1*0, 1*(- 1) - 2*2]

= [- 1, -2, -5]

The area of the triangle ABC = 0.5|AB||AC|sin (angle BAC).

|AB| is the modulus of vector AB, that is, the length of vector AB.

= 0.5*| Cross product of vector AB and vector AC |

= 0.5*|[- 1,-2,-5]|

= 0.5*[ 1^2 + 2^2 + 5^2]^( 1/2)

= 0.5*[30]^( 1/2)

|AD|^2 = 2^2 + 2^2 + 1^2 = 9，|AD| = 3。

|bd|^2 = 1^2+0^2+0^2 = 1，|BD| = 1。

|cd|^2 = 0^2+3^2+ 1^2 = 10，|CD| = ( 10)^( 1/2)

The distance from triangle ABC to point D (3, 1, 1) = the minimum distance from three vertices of triangle ABC to point D (3, 1, 1).

= |BD| = 1