The resultant of the first two vectors is 4i+(5b+3)j+6k; and the second two: -3i+3bj+k, assuming (2b-i) should be (2b-1).
The two resultants are orthogonal which means their dot product (scalar product)=0:
-12+3b(5b+3)+6=0; 15b^2+9b-6=0; 5b^2+3b-2=0=(5b-2)(b+1), so b=-1 or 2/5.