Solving for Lambda and Mu

We would have to solve part a and b first in order to answer part c. 

a)  $OX=\lambda OY =\lambda (a+2b)$

b) $OX=OQ+QX=3b+\mu QP=3b+\mu (a-3b)$

c) From parts a and b we have
\[OX=\lambda (a+2b)=3b+\mu (a-3b).\]
Thus 
\[\lambda a+2\lambda b=3b+\mu a-3\mu b\]
\[\Rightarrow \lambda a- \mu a =3b -3\mu b-2\lambda b\]
\[\Rightarrow   (\lambda -\mu ) a =(3 -3\mu -2\lambda) b,\]
sinc the vectors $a$ and $b$ are not parallel, we must have
\[(\lambda -\mu )=0   \text{and}  (3 -3\mu -2\lambda)=0.\]
Hence $\lambda=\mu$, and subsituting in the second equation above we get 
\[3 -3\mu -2\lambda=3- 3 \lambda -2\lambda=0  \Rightarrow 3-5\lambda=0\]
\[\Rightarrow \lambda =\frac{3}{5}.\]

Hence $\lambda=\mu=\frac{3}{5}$.

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Note that if $\lambda -\mu $ and $3 -3\mu -2\lambda$ are not zero, then 
\[a=\frac{3 -3\mu -2\lambda}{\lambda -\mu}b\]
which means that the vectors $a$ and $b$ are parallel, which is a contradiction.