Given,
x+2y+3z=42,x,y,z≥0
Now, taking cases for different value of z we get,
z=0; x+2y=42
Now, y can take values 0,1,2,....21→22elements
Similarly, we get
z=1; x+2y=39⇒20elements
z=2; x+2y=36⇒19elements
z=3; x+2y=33⇒17elements
z=4; x+2y=30⇒16elements
z=5; x+2y=27⇒14elements
z=6; x+2y=24⇒13elements
z=7; x+2y=21⇒11elements
z=8; x+2y=18⇒10elements
z=9; x+2y=15⇒8elements
z=10; x+2y=12⇒7elements
z=11; x+2y=9⇒5elements
z=12; x+2y=6⇒4elements
z=13; x+2y=3⇒2elements
z=14; x+2y=0⇒1elements
Total : 169