Solution verification: sum at least $4N/7$ times odd 
Problem: Let $a_{j}$,$b_{j}$,$c_{j}$ be whole numbers for $ 1 \leq j \leq N$. Suppose that for each $j$ at least one of $a_{j}$,$b_{j}$,$c_{j}$ is odd. Show that there are whole numbers $r$,$s$ and $t$ such that the sum $$r\cdot a_{j} + s\cdot b_{j} + t\cdot c_{j}$$ (for $ 1 \leq j \leq N$) is in at least $4N/7$ cases odd.

My attempt: Choose $r=s=t=1$. To get an odd sum with 3 numbers for which one is odd for sure you must have that: 2 are even and the other odd or all three are odd. The total amount of possible even-odd combinations for 3 numbers is equal to ${3 \choose 1} +{3 \choose 2} +{3 \choose 3}=7$ (1 odd v 2 odd v 3 odd), so with $r=s=t=1$ we see that in $4/7$ cases, the sum is odd for a fixed $j$. In total we have thus that in $4N/7$ circumstances the sum is odd.
My doubts: So, while reading the question I thought this would seem like a classic pigeonhole principle question and at first I had the intention to go that route, but when I was trying some random cases I came across this with $r=s=t=1$ and with this all the restrictions hold. What do you guys think about my solution?

This does not work. When all $a_j, b_j$ are odd and all $c_j$ are even, $r=s=t=1$ never works. You can solve this problem by considering multiple possibilities for $r,s,t$ and showing at least one will work.
Hint: There are $8$ possible parities for $(r,s,t).$
