Vector trapezium help

I have no idea
Have tried everything, can't find any similar questions (they always give a ratio)

You could represent a point lying on OC as
m*(3a+4b)
Where m is between 0 and 1.

Do the same for a point on the other line in terms of n, then equate and solve for m and/or n and hence get the intersection point.

I'm not trying to be rude but could you solve it for me
I've been stuck on this question all day and even asked in 5 discord servers for help and nobody has been able to solve it

Have you done the first part, representing a point on AB in terms of a,b,n?

OA = 3a
OC = 3a+4b = k(ON)
ON = k(3a+4b)
OB = 6b
AB = 6b - 3a
this is as far as I can go

OC = 3a+4b
So a point P lying on that is
mOC = m(3a+4b)

OA = 3a
AB = 6b-3a
So a point P lying on there is
OA + nAB = ...

Then equate and solve for m and n, noting that equality must hold for both directions a and b.

I understand this but how do you make an equation out of this? I inputted 6b-3a=0 and 3a+4b=0 on wolfram alpha in the system of equations calculator but it didn't give me an answer oK

I have no idea what you tried to do on wolframalpha.

if you're following above advice you have the equations for point(s) on two lines in terms of m and n. At intersection, they must be equal so equate and solve for m and n.

And I'm not being funny by saying I'm not doing it for you. You should sketch what it means, think about how to solve it. Take more time than a minute or two before hitting reply. All the relevant advice is in this, and previous, posts.

What did you do? Did you equate the point equations, how did you try and solve them, ...?

A quick Google gave a related problem
https://www.bbc.co.uk/bitesize/guides/zy8d6yc/revision/3
Two lines share the same point so they set them equal to each other. There are other, similar examples.

In this problem you can think of a and b vectors as being your coordinate axes, but they're not perpendicular to each other (not really important).