Real numbers a, b, c and d satisfy:

(b+a^2-3lna)^2+(c-d+2)^2=0

There are:

B+A 2-3 LNA = 0, let b=y and a=x, then there is: Y = 3LNX-X 2.

C-d+2=0, let c=x, d=y, then there is: y=x+2.

So:

(a-c) 2+(b-d) 2 is the square of the minimum distance between the curve y = 3lnx-x 2 and the straight line y=x+2.

Derive the curve y = 3lnx-x 2:

y'(x)=3/x-2x

The slope of the tangent parallel to y=x+2 is k= 1=3/x-2x.

Solution: x = 1 (x =-3/2

X= 1 substitute y = 3lnx-x 2 to get: y=- 1.

So: the tangent point is (1,-1).

Distance from tangent point to straight line y=x+2:

l=| 1+ 1+2|/√( 1^2+ 1^2)=2√2

So: l 2 = 8

So the minimum value of (a-c) 2+(b-d) 2 is 8.

Option d