Find a vector parallel to the line of intersection of the planes 2x−2y−3z=7 and −x+4y−5z=0.

The normal to the given plane P1: 2x−2y−3z=7 is N1=<2,-2,-3>, and similarlythe normal to the given plane P2: −x+4y−5z is N2=<-1,4,-5>, and similarly
The normal to the above normals is parallel to the intersection of planes P1 and P2, which is given by the cross product of N1 and N2:V = N1 x N2 =    i    j    k   2  -2  -3  -1   4  -5= V<10+12, 3+10, 8-2 >=V<22, 13, 6> 
 
A vector parallel to the intersection of P1 and P2 is V<11, 7, 3>.
Check: substitute V in P1: 2(22)-2(13)-3(6)=0 => V is parallel to P1.substitute V in P2: -(22)+4(13)-5(6)=0 => V is parallel to P2Therefore V is parallel to intersection of P1 and P2.
 Read more on - link