Let M(1; -1; 0) be a point, Delta: (x-2)/2 =(y+1)/(-1) = (z - 1)/1 be a line, (P): x + y + z - 2 = 0 be a plane. Find the coordinates of the point A lies on (P), knowing that the line AM perpendicular to the line Delta and distance from the point A to the line Delta equal to sqrt(33/2).

This is my code

 with(geom3d):

point(M,1,-1,3):

line(Delta,[2*t+2,-t-1,t+1],t):

plane(P,x+y+z+3=0,[x,y,z]):

a:=ParallelVector(Delta):

Let (d1): x = 2*t +1, y = t-2, z =-2*t+2 and 

(d2): x = 2-m, y = 3+m,z = 4+m 

be two lines and (P): x - y + z - 6 = 0 be a plane. Write the equation of the line (Delta), knowing that (Delta) parallel to (P) and cuts (d1), (d2) at the points M and N, respectively so that MN = 3*sqrt(6).

Let A(1, -1, 3) is a point; (P): x + y + 5z = 0 be a plane; (l): x= 2t, y = -5t, z = 6t be a line. Write the equation of a line Delta parallel to the plane (P) and has a distance to (l) equal to 1.

I was thinking about this problem many times but i can not 

Problem. Let A(-1,5) and B(-2, -2) be two points and l: 3*x - 4*y -27 = 0. Write the equation of the circle passing through points A, B and tangent to the line l.

This is my code.

Restart: with(geometry):

point(A,-1,5):

point(B,-2,-2):

point(C,5,-3):

circle(ABC,[A,B,C],[x,y]):

sort(Equation(ABC)):

tangentpc(l, ABC,C):

sort(Equation(l)):

point(T,a,b):sys:=solve([distance(T,A) = distance(T,B), distance(T,A)=distance(T,l)],[a,b]):

Let A(-5,-3,-3), B(0,1,-2) be two points and

(d): x = t - 3, y = 2*t, z = t+2.

Find the coordinates the point M on (d) so that the area of the triangle ABM obtain minimum value.

This is my code.

restart; with(LinearAlgebra):

A:=: B:=: M:=:

u:=A-B: v:=A-M:

T:=CrossProduct(u,v):

S:=minimize(1/2*Norm(T,2), location = 'true');

M:=subs(op(1,op(1, S[2])),M);