Python 2, 203 200 bytes

E=enumerate
l=input()
t=eval(`[[0]*-~len(l)]*4`)
for i in range(3600):
 for y,q in E(t):q[sum(y==((sum([~ord(w)%5*[j%3]+[3]for j,w in E(c)],[])+[3]*6)*450)[i]for c in l)^0-y/3]+=y*y|5>t[2][-1]
print t

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Python 2, 203 200 bytes

E=enumerate
l=input()
t=eval(`[[0]*-~len(l)]*4`)
for i in range(3600):
 for y,q in E(t):q[sum(y==((sum([~ord(w)%5*[j%3]+[3]for j,w in E(c)],[])+[3]*6)*450)[i]for c in l)^0-y/3]+=y*y|5>t[2][-1]
print t

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Building the pattern with strings instead of lists of digits comes out at the same length and is a bit faster: Try it online!

Python 2, 203 bytes

R=range
l=input()
t=eval(`[[0]*-~len(l)]*4`)
for i in R(3600):
 for y in R(4-t[2][-1]/5*3):t[y-1][sum(y==((sum([~ord(c[j])%5*[j%3+1]+[0]for j in R(len(c))],[])+[0]*6)*450)[i]for c in l)^-0**y]+=1
print t

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Python 2, 203 200 bytes

E=enumerate
l=input()
t=eval(`[[0]*-~len(l)]*4`)
for i in range(3600):
 for y,q in E(t):q[sum(y==((sum([~ord(w)%5*[j%3]+[3]for j,w in E(c)],[])+[3]*6)*450)[i]for c in l)^0-y/3]+=y*y|5>t[2][-1]
print t

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Python 2, 205 bytes

l=input()
t=eval(`[[0]*-~len(l)]*4`)
for i in range(3600):
 for y in range(4-t[2][-1]/5*3):t[y-1][sum(y==((sum([~ord(w)%5*[j%3+1]+[0]for j,w in enumerate(c)],[])+[0]*6)*450)[i]for c in l)^-0**y]+=1
print t

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Python 2, 203 bytes

R=range
l=input()
t=eval(`[[0]*-~len(l)]*4`)
for i in R(3600):
 for y in R(4-t[2][-1]/5*3):t[y-1][sum(y==((sum([~ord(c[j])%5*[j%3+1]+[0]for j in R(len(c))],[])+[0]*6)*450)[i]for c in l)^-0**y]+=1
print t

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