1. We need the number of non-negative integer solutions of $x+2y+3z=42,$ where $x,y,z\in \mathbb Z$ and $x,y,z\ge 0$.

2. For each fixed $z$, the equation becomes $x+2y=42-3z.$ Since $x,y\ge 0$, we must have $42-3z\ge 0 \implies z=0,1,2,\dots,14.$

3. Now count the number of non-negative integer solutions of $x+2y=N$ for each $N=42-3z$.

For a fixed $N$, once $y$ is chosen, $x=N-2y$ is determined and must be non-negative. So $y=0,1,2,\dots,\left\lfloor \frac N2 \right\rfloor.$ Hence number of solutions is $\left\lfloor \frac N2 \right\rfloor+1.$

Thus total number of solutions is $\sum_{z=0}^{14} \left(\left\lfloor \frac{42-3z}{2} \right\rfloor+1\right).$

4. Evaluate this sum term-by-term:

• $z=0$: $N=42$, count $=\lfloor 42/2\rfloor+1=21+1=22$
• $z=1$: $N=39$, count $=\lfloor 39/2\rfloor+1=19+1=20$
• $z=2$: $N=36$, count $=18+1=19$
• $z=3$: $N=33$, count $=16+1=17$
• $z=4$: $N=30$, count $=15+1=16$
• $z=5$: $N=27$, count $=13+1=14$
• $z=6$: $N=24$, count $=12+1=13$
• $z=7$: $N=21$, count $=10+1=11$
• $z=8$: $N=18$, count $=9+1=10$
• $z=9$: $N=15$, count $=7+1=8$
• $z=10$: $N=12$, count $=6+1=7$
• $z=11$: $N=9$, count $=4+1=5$
• $z=12$: $N=6$, count $=3+1=4$
• $z=13$: $N=3$, count $=1+1=2$
• $z=14$: $N=0$, count $=0+1=1$

5. Add them: $22+20+19+17+16+14+13+11+10+8+7+5+4+2+1.$

Let's sum carefully: $22+20=42$ $42+19=61$ $61+17=78$ $78+16=94$ $94+14=108$ $108+13=121$ $121+11=132$ $132+10=142$ $142+8=150$ $150+7=157$ $157+5=162$ $162+4=166$ $166+2=168$ $168+1=169.$

Therefore, the number of elements in the set is $\boxed{169}.$