C4 vectors help!

I am doing this vectors question an der I am really stuck on 1c)! Please help
http://www.southwolds.co.uk/user/59/49197.pdf
The answer to 1a) is R = (3,4,6). 1c): q= 2 or -4
Many thanks ☺

You know P, you know R, and you can express Q as $(1-2q,5+q,6)$

The triangle's area is given by $\frac{1}{2} |\overrightarrow{RP}| |\overrightarrow{RQ}|$ (since it's right-angled). Evaluate it in terms of $q$, set it equal to $9\sqrt{5}$ and solve the quadratic for the two values of $q$

Thanks for replying!
I've done that but I keep getting an answer of 3?!
I got the length of PR to be 6 so the length of RQ should be 3root5.
The length/ modulus of the direction vector is root5, suggesting q is 3. I think I need to use (1,5,6) in some way but I'm not sure how.

$| \overrightarrow{PR} | = 6$ - correct.

$| \overrightarrow{RQ} | = 3\sqrt{5}$ - not sure where you are pulling this from.

As I said, we know R and we can express Q in terms of $q$. Hence $\overrightarrow{RQ} = (-2-2q, 1+q, 0)$ which has magnitude....?

1/2 x 6 x mod RQ = 9 root 5
If you divide by 3 you get 3 root 5 for mod RQ.
I've got that the magnitude of RQ = root(5q^2 + 10q + 5) so =q+1? But this doesn't give you the value of q

$5q^2+10q+5=5(q+1)^2$ so $|\overrightarrow{PQ}| = |q+1|\sqrt{5}$

Hence $|q+1| \sqrt{5} = 3 \sqrt{5}$

Solve for the two values of q.