Use awk command to get information below a pattern - linux

Question

I have a file with a wide range of information and I want to extract some data from here. I only will post here the interesting part. I want to extract IQ and JQ values as well as the J_ij[meV] value which is two lines above. I read this question How to print 5 consecutive lines after a pattern in file using awk where a pattern is used to extract information bellow and I was thinking doing something similar. My initial idea was:
awk '/IQ =/ { print $6,$12 } /IQ =/ {for(i=2; i<=2; i++){ getline; print $11 }}' input.text > output.txt
Loop appears not to working
IT IQ JT JQ N1 N2 N3 DRX DRY DRZ DR J_ij [mRy] J_ij [meV]
IT = 1 IQ = **1** JT = 1 JQ = **1**
->Q = ( -0.250, 0.722, 0.203) ->Q = ( -0.250, 0.722, 0.203)
1 1 1 1 0 0 0 0.000 0.000 0.000 0.000 0.000000000 **0.000000000**
IT = 1 IQ = **1** JT = 6 JQ = **6**
->Q = ( -0.250, 0.722, 0.203) ->Q = ( 0.000, 1.443, 0.609)
1 1 6 6 -1 0 -1 -0.250 -0.144 -0.406 0.498 0.135692822 **1.846194885**
IT = 1 IQ = **1** JT = 8 JQ = **8**
->Q = ( -0.250, 0.722, 0.203) ->Q = ( 0.000, 0.577, 0.609)
1 1 8 8 0 0 -1 0.250 -0.144 -0.406 0.498 0.017676555 **0.240501782**
My expected output is:
IQ JQ J_ij [meV]
1 1 0.000000000
1 6 1.846194885
1 8 0.240501782
It comes from the bold words (** **), first line is only indicative.

Answer 1

Could you please try following. Written and tested with shown examples.
awk '
BEGIN{
print "IQ JQ J_ij [meV]"
}
FNR>1 && /IQ =/{
value=$6 OFS $12
found=1
next
}
found && NF && !/ ->Q/{
if(value){
print value OFS $NF
}
value=found=""
}' Input_file
Output will be as follows.
IQ JQ J_ij [meV]
1 1 0.000000000
1 6 1.846194885
1 8 0.240501782