The correct option is B 55
Let r=xi+yj+zk
Thus, r.a=x+y+z
Hence, x+y+z=12
=>x+y+z=12
This is thus equivalent to distributing 12 identical sweets among 3 boys.
Now, since x, y, z are positive integers, we give them 1 sweet each. Thus, we are left with 12−3=9
Thus, we need to distribute 9 sweets among 3 boys where a boy getting no sweet is also allowed.
This can be done in9+3−1C3−1=11C2=55 ways.
Hence, (b) is correct.