I want to put one option of tetrahedron SABC with AB=c, BC=a, AC=b, SA=m, SB=n, SC=p. I tried
restart;
with(Student:-MultivariateCalculus);
A := [0, 0, 0];
B := [c, 0, 0];
C := [(-a^2 + b^2 + c^2)/(2*c), sqrt((a + b - c)*(a - b + c)*(-a + b + c)*(a + b + c))/(2*c), 0];
S := [x, y, z];
solve([distance(S, A) = m, distance(S, B) = n, distance(S, C) = p], [x, y, z]);

I got warning, solutions may have been lost. How can I solve that system of equations?

I am trying to find remainder of (2 + sqrt(3))^15 + (2 - sqrt(3))^15 with 2017. I tried
irem((2 + sqrt(3))^15 + (2 - sqrt(3))^15, 2017);

I do not get the result. How can I get the result?

I use like this 
a := expand((2 + sqrt(3))^15 + (2 - sqrt(3))^15);
irem(a, 2017);

get the result.

How do I draw this polyhedron knowing all vertices?
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_21?fbclid=IwAR2bdHwJEpjarZEwope1hYOcnnXqnvzae38Y8ZrfWKCKQunfAuT9s30q32o

[[-1.5, -0.5, 0.], [-1.5, 0.5, 0.], [-1., -1., -0.707107], [-1., 1., -0.707107], [-0.5, -1.5, 0.], [-0.5, -0.5, -1.41421], [-0.5, 0.5, -1.41421], [-0.5, 1.5, 0.], [0.5, -1.5, 0.], [0.5, -0.5, -1.41421], [0.5, 0.5, -1.41421], [0.5, 1.5, 0.], [1., -1., -0.707107], [1., 1., -0.707107], [1.5, -0.5, 0.], [1.5, 0.5, 0.]]

The date, month, year  22/02/2022 is said a beautiful numbers. Because, the number 2202 symmetry to the number 2022. How can I find another beautiful numbers like that ?

With two vectors a and b, we know that
Norm(CrossProduct(a, b)) = Norm(a)* Norm(b) * sin(a,b).
I tried with a := <1, 2, -2>; b := <2, 10, 11>; 

Note that a perpendicular to b and 

Norm(CrossProduct(a, b)) = Norm(a)* Norm(b)

I tried

restart;
with(VectorCalculus);
SetCoordinates(cartesian[x, y, z]);
a := <1, 2, -2>;
b := <2, 10, 11>;
Norm(a);
Norm(b);
v := CrossProduct(a, b);
Norm(v);

Are there two vectors a and b with integer coordinates and  not perpendicular,  so that Norm(a), Norm(b), Norm(CrossProduct(a, b)) are interger numbers satisfying
Norm(CrossProduct(a, b)) = Norm(a)* Norm(b) * sin(a,b).