Q:

In Bash, grep for a line and print that line in a file but with more information

In bash, I am using the grep command to print the result to a file:
  grep "4 CA   1" CVOLOPTs*SYMREMO.out | grep "11 O    0 0 0" >> data.dat

Which prints the following in the data.dat file:
 4 CA   1     2.3311     4.4052   11 O    0 0 0

I would like to make the data.dat look like the following:
 252   4 CA   1     2.3311     4.4052   11 O    0 0 0

I have tried something like:
grep "4 CA   1" CVOLOPTs*SYMREMO.out | grep "11 O    0 0 0" >> echo "252" data.dat

But does not work.
I would appreciate if you could help me please.

A:

An additional pipe with awk could do the trick:
grep "4 CA   1" CVOLOPTs*SYMREMO.out | grep "11 O    0 0 0" | awk ´{print "252",$0}´ >> data.dat

the whole thing can be done with a single awk:
awk ´/4 CA   1/ && /11 O    0 0 0/ {print "252",$0}' CVOLOPTs*SYMREMO.out >> data.dat

$0 is the entire current line. In this situation, the line selected by the 2 regex.
awk reads each line one by one.
Regex stands for regular expression. With awk they are enclosed into /:  e.g. /4 CA   1/
/4 CA   1/ && /11 O    0 0 0/ is an expression: it is true when the line being read matches both the first regex and (&&) the second regex. In this situation there are just strings; see 
9. Regular Expressions from the Open Group Base Specifications Issue 6.

