;
I first rather lazily tried Magma's command
EliminationIdeal(I,2);
The result was a degree-55 polynomial in x3, which factors as
x3^3*(quartic)^3*(irreducible degree-40).
So we expect maybe 45 or so points in the intersection. But that's missing
the "exceptional" points (where several have the same x3).
So really we have to Do The Right Thing. The question here is the number of
components in a variety of (we hope) dimension zero, determined by an
ideal generated by three polynomials in (say) Z[x1, x2, x3]. The algebraic
geometry involved can be expressed ring-theoretically: we need to know
the decomposition of the radical of that ideal as an intersection of
prime ideals. High-flying talk? Fear not. Magma also has a command
RadicalDecomposition(I);
to replace the EliminationIdeal. It computes exactly these primes, and
as a bonus determines the number of points in each of the corresponding
varieties. With a few more hours of machine time we find the variety
has four points with x3=0, four with x2=0, and 40 in an irreducible component
with neither x2=0 nor x3=0.
(Actually I got this result faster first and then corroborated more
slowly: Declaring instead p:=157526249; (or some other large prime) and
K:=GF(p); gives the decomposition into prime ideals fairly quickly; one
notices the x2=0 and x3=0 patterns quickly, and verifies them integrally,
and then finds the remaining factor must be irreducible to be consistent with
the data from a few such primes. Of course, it _could_ happen that I just
happened to choose primes from among the set over which the number of
points decreases from the integral picture, so I had to verify the whole
calculation integrally, which took much longer. OK, I'm lying: I killed it
after about an hour.)
Since we're going more carefully here, it's worth noting that the cases
x2=0 or x3=0 don't translate so nicely to the original 5-variable problem.
One can re-run the program starting with a 5-variable ring, and do it
all properly. Each of the 40 points indicated above does indeed lead to
a (unique) point in C^5; those with x2=0 or x3=0 are not in the image
of the variety in C^5.
So, after many wasted machine cycles, we are led to the conclusion
that there are 40 points in this solution set in C^5.
Of course, this all pertains only to the _particular_ choices I made for
L2 L5 L7 S1 S2, and assumes L1, L3, and L6 are linearly independent.
I tried another set of choices:
p:=10000000019;
K:=GF(p);
P__:=PolynomialRing(K,5,"lex");
L1:=x1;L4:=x2;L6:=x3;
c0:=1; c1:=1; c2:=2; c3:=13;
L2:=c0+c1*x1+c2*x2+c3*x3;
L7:=1-4*x1+10*x2+16*x3;
L5:=4-x1-8*x2+6*x3;
S1:=2+x1-x2+3*x3-11*x1^2+2*x2^2+5*x3^2-6*x1*x2+7*x1*x3-17*x2*x3;
S2:=+1-2*x1+99*x2+0*x3+7*x1^2-111*x2^2+13*x3^2-17*x1*x2+11*x1*x3-23*x2*x3;
I:=ideal__

;
RadicalDecomposition(I);
Conclusion: another set of 40 points mod p, presumably 40 points over
the rationals too. Taking another leap of faith, we expect 40 points
in C^5 for almost all choices of the L's and S's.
(I wanted to be a little more general, so I tried replacing the first Magma
ring (and leaving c0...c3 variable):
L:=GF(p); K