How to find the mass spectrum of Cl3+?
Okay, let's tackle this. First of all you worked out correctly that there are four different combinations, resulting in four different peaks. You have also given the probability for finding a $\ce{{}^{35}_{17}Cl}$ is $a=33.3\%\approx\frac13$ and the probability for finding a $\ce{{}^{37}_{17}Cl}$ is $b=66.7\%\approx\frac23$. Since you have three positions to fill, this will result in a binomial formula of third order, $(a+b)^3$. You can solve this easily with Pascal's triangle, $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. You spectrum should therefore look like this table:
\begin{array}{lrr}\hline
   &amp; \text{m/z} &amp;\text{abundance}/\%\\\hline
\ce{[{}^{35}_{17}Cl_3]^{+}}                &amp; 105   &amp;  3.7  \\
\ce{[{}^{35}_{17}Cl_2{}^{37}_{17}Cl]^{+}}  &amp; 107   &amp; 22.2  \\
\ce{[{}^{35}_{17}Cl{}^{37}_{17}Cl_2]^{+}}  &amp; 109   &amp; 44.4  \\
\ce{[{}^{37}_{17}Cl_3]^{+}}                &amp; 111   &amp; 29.7  \\\hline
\end{array}

The following snowball diagram should further demonstrate this: