show that there are $1\leq j_1< j_2< j_3\leq 13$ such that $\left | A_{j1}\cap A_{j2}\cap A_{j3} \right |\geq 3$. Given $A_i,A_2,...,A_{13}$ $\subset [10]$ when $\left | A_i \right |=5$ for all $i$.
Need to show that there are $1\leq j_1< j_2< j_3\leq 13$ such that $\left | A_{j_1}\cap A_{j_2}\cap A_{j_3} \right |\geq$ $3$.
I tried to put the numbers but didn't get how to prove.
Thanks.

Your claim is false as currently stated (though I think
it becomes true if $13$ is replaced by $14$). Here is a 
counterexample :
$$
\begin{array}{lcllcllcl}
A_1 &=& \lbrace 1,2,3,4,5 \rbrace, &
A_2 &=& \lbrace 1,2,3,4,6 \rbrace, &
A_3 &=& \lbrace 1,2,5,6,7 \rbrace, \\
A_4 &=& \lbrace 1,2,7,8,9 \rbrace, &
A_5 &=& \lbrace 1,3,5,6,7 \rbrace, &
A_6 &=& \lbrace 1,3,7,8,10 \rbrace, \\
A_7 &=& \lbrace 1,4,5,9,10 \rbrace, &
A_8 &=& \lbrace 2,3,5,6,8 \rbrace, &
A_9 &=& \lbrace 2,3,7,8,10 \rbrace, \\
A_{10} &=& \lbrace 2,4,5,7,9 \rbrace, &
A_{11} &=& \lbrace 2,4,6,7,10 \rbrace, &
A_{12} &=& \lbrace 3,4,6,7,9 \rbrace, \\
A_{13} &=& \lbrace 4,5,6,8,10 \rbrace. & & & & & \\
\end{array}
$$
