Question: Coordinates of a point (3)

Let A(1,0,0) be a point and two lines

d1: (x-5)/3 = (y-1)/1 = (z - 2)/2,

d2: (x-5)/1 = (y-1)/1 = (z - 3)/3

Let B and C be two point lies on d1 and d2, respectively so that the three points A, B, C lie on a straight line. Find the coordinates   the point M on the axis Oy so that the area of the triangle ABC equal to 3.

This is my code

restart:
A:=<1,0,0>:  B:=<3*t+5,1 + t,2*t+2>: C:=<5+m, 1+m,3+3*m>:
u:=A - B: v:=A - C:
w:=LinearAlgebra[CrossProduct](u,v): solve([seq(w[i]=0,i=1..3)]): assign(%):
'B'=B; 'C'=C; M:=<0,a,0>:
with(LinearAlgebra):
T:=CrossProduct(B-M,B-C):
sol:=solve(1/2*Norm(T,2)=3,{a});
for i from 1 to 2 do print('M'=(M[1], op(2,op(sol[i])),M[3])) end do;

Please comment to me.  Especially, the line

for i from 1 to 2 do print('M'=(M[1], op(2,op(sol[i])),M[3])) end do;

I can not use 

for i to nops(sol) do print('M'=(M[1], op(2,op(sol[i])),M[3])) end do;

Please help me. Thank you very much.