Let S1: x^2 + (y - 2)^2 + (z + 1)^2 = 29 be a sphere and two points A(0, 0, 4), B(6, -2, 6); the line d passing through point C(4, -8, 4) and have direction v=(1, -1, 2). Find the point M such that M lies on the sphere S1, the angle AMB equals to 90 degree and distance from M to the line d is minimum. 

This is my code
 

restart;
with(Student:-MultivariateCalculus);
A := [0, 0, 0];
B := [a, 0, 0];
C := [a, a, 0];
DD := [0, a, 0];
S := [0, 0, h];
d1 := Line(A, C);
d2 := Line(S, DD);
(Distance(d1, d2) assuming (a::positive and h::positive));

I got 

abs(a)^2*abs(h)/sqrt(2*abs(a*h)^2 + abs(a)^4)

How can I remove abs?

How can I find a relation of a, b, c, d, t (a d <> 0) so that the function f(x) = (a x^2 + b x + c)/(d x +t) is a increasing function on the interval (m, n)? I know that, we have the conditions 
f'(x) >=0 forall x in (m, n) and -t/d not in (m, n). 

Let SABCD be a pyramid, with the line SA perpendicular to the plane ABC, ABCD is a rectangle, AB = a, AB = b, SA = h; H is projection point of A on the line SC and K is projection point of A on the line SD. How can I find radius of the circumcircle of triangle CHK?

restart;
with(Student:-MultivariateCalculus);
A := [0, 0, 0];
B := [a, 0, 0];
C := [a, b, 0];
DD := [0, b, 0];
S := [0, 0, h];
lineSC := Line(S, C);
lineSD := Line(S, DD);
H := Projection(A, lineSC);
K := Projection(A, lineSD);

My code
restart;
reqn:=f(n+1) = f(n)  + sqrt(2*f(n+1)-f(n)):
rsolve({reqn, f(1) = 15},f(n));
I can not get the answer. How can I get the answer?