Answer: \(\mathbf{v}_A=-\frac52\mathbf{i}-6\mathbf{j}\), speed of \(B=15\text{ m s}^{-1}\), collision at \(t=6\) and position vector \(5\mathbf{i}-43\mathbf{j}\).

The direction vector for \(A\) is \(-5\mathbf{i}-12\mathbf{j}\). Its magnitude is

\(\sqrt{(-5)^2+(-12)^2}=13.\)

So the unit vector in this direction is

\(-\frac5{13}\mathbf{i}-\frac{12}{13}\mathbf{j}.\)

Since the speed is \(6.5=\frac{13}{2}\),

\(\mathbf{v}_A=\frac{13}{2}\left(-\frac5{13}\mathbf{i}-\frac{12}{13}\mathbf{j}\right) =-\frac52\mathbf{i}-6\mathbf{j}.\)

The speed of \(B\) is

\(|\mathbf{v}_B|=\sqrt{12^2+(-9)^2}=\sqrt{225}=15.\)

The position vector of \(A\) after \(t\) seconds is

\(\mathbf{r}_A=(20\mathbf{i}-7\mathbf{j})+t\left(-\frac52\mathbf{i}-6\mathbf{j}\right) =\left(20-\frac52t\right)\mathbf{i}+(-7-6t)\mathbf{j}.\)

The position vector of \(B\) after \(t\) seconds is

\(\mathbf{r}_B=(-67\mathbf{i}+11\mathbf{j})+t(12\mathbf{i}-9\mathbf{j}) =(-67+12t)\mathbf{i}+(11-9t)\mathbf{j}.\)

At collision, \(\mathbf{r}_A=\mathbf{r}_B\). Comparing the \(\mathbf{i}\)-components,

\(20-\frac52t=-67+12t.\)

So

\(87=\frac{29}{2}t,\)

and hence

\(t=6.\)

Substituting into \(\mathbf{r}_A\),

\(\mathbf{r}=5\mathbf{i}-43\mathbf{j}.\)

So the particles collide after \(6\) seconds at position vector

\(\boxed{5\mathbf{i}-43\mathbf{j}}.\)