@Carl Love

If you want to see the gridlines on the filled region then  transparency  option can be used. 

@Carl Love

If you want to see the gridlines on the filled region then  transparency  option can be used. 

According to Markiyan's idea:

l1:=[a1*t+x1, b1*t+y1, c1*t+z1]:

l2:=[a2*s+x2, b2*s+y2, c2*s+z2]:

minimize(add((l1[i]-l2[i])^2, i=1..3), s=-infinity..infinity, t=-infinity..infinity) assuming a1^2+b1^2+c1^2>0 and a2^2+b2^2+c2^2>0:

RealDomain[simplify](sqrt(%)); 

@Markiyan Hirnyk 

Let the questioner will decide himself which option is more suitable for him. Purely terminological distinction.

@Markiyan Hirnyk 

Let the questioner will decide himself which option is more suitable for him. Purely terminological distinction.

@Markiyan Hirnyk 
The questioner wrote "two positive solutions and two negative solutions". If  m=5/2  then we have one negative and two positive roots.

@Markiyan Hirnyk 
The questioner wrote "two positive solutions and two negative solutions". If  m=5/2  then we have one negative and two positive roots.

@Markiyan Hirnyk 

Of course, I understand that the field is not potential, so the path of integration should be specified.

@Markiyan Hirnyk 

Of course, I understand that the field is not potential, so the path of integration should be specified.

This is a simple task, it is easy to solve by a procedure. But I did not understand the law of getting your matrix for any m.  What is  i ? Can you give an example of such a matrix for a given m, for example, for m = 8 ?

I noticed an error in the text of the procedure. Instead of  if alpha = 0 then {k> 0}  should be  if alpha = 0 then print(`No solutions`) .

@Markiyan Hirnyk

Of course, the advantage of substantially purely aesthetic.
Also, I think this form of writing can be interesting for educational purposes in the study of the concept of absolute value.

PS. The advantage is the same as   |x|  over 

I did not notice the condition that  x>=0, y>=0, z>=0. Therefore, the solution exists.

I did not notice the condition that  x>=0, y>=0, z>=0. Therefore, the solution exists.

@one_man

Thank you for your interest! I was waiting for your question, because I know about your enthusiasm implicitly given curves. Setting the boundary of the figure is not provided in this way in my procedures, but still there is a simple solution to the problem. Here it is: 

P1 := plots[implicitplot]((x1+0.7e-1*sin(30*x2)^2)^2+(x2+0.7e-1*sin(30*x1)^2)^2-1 = 0, x1 = -2 .. 2, x2 = -2 .. 2, numpoints = 100000):

P2 := plottools[getdata](P1):

L := [convert(P2[3], listlist)]:

Perimeter(L);

Area(L);

plots[display](Picture(L, color = green, [color = black, thickness = 2]));