Answer: \(\begin{pmatrix}36\\-15\end{pmatrix}\); \(\lambda=-\frac{15}{22}\), \(\mu=-\frac{20}{11}\).

(a) The magnitude of \(\begin{pmatrix}12\\-5\end{pmatrix}\) is

\(\sqrt{12^2+(-5)^2}=\sqrt{169}=13.\)

To make the magnitude \(39\), multiply the vector by \(3\):

(b) We have

and

Also

Equating components gives

\(10+4\lambda=-4\mu\)

and

\(-5+6\lambda=5\mu.\)

From the first equation, \(\mu=-\frac52-\lambda\). Substituting into the second equation gives

\(-5+6\lambda=5\left(-\frac52-\lambda\right).\)

So

\(11\lambda=-\frac{15}{2},\)

hence

\(\lambda=-\frac{15}{22}.\)

Then

\(\mu=-\frac52+\frac{15}{22}=-\frac{20}{11}.\)