Use two line parameters, a division ratio and a cross-product area

On BC,

$\begin{aligned}&\mathbf r\\ & =\mathbf b+\lambda(3\mathbf a-2\mathbf b)\\ & =3\lambda\mathbf a+(1-2\lambda)\mathbf b.\end{aligned}$

On AD,

$\begin{aligned}&\mathbf r\\ & =\mathbf a+\mu(\mathbf a+3\mathbf b)\\ & =(1+\mu)\mathbf a+3\mu\mathbf b.\end{aligned}$

Since a,b are nonparallel in the diagram, compare coefficients: $3\lambda-\mu=1$, $2\lambda+3\mu=1$. Thus $\lambda=4/11$, $\mu=1/11$ and

$OX=\frac{12}{11}\mathbf a+\frac 3{11}\mathbf b.$

Use the source alternative parameters $AX:XD=\lambda:(1-\lambda)$ and $BX:XC=\mu:(1-\mu)$. The section formula gives

$\begin{aligned}&OX\\ & =(1+\lambda)\mathbf a+3\lambda\mathbf b\\ & =3\mu\mathbf a+(1-2\mu)\mathbf b.\end{aligned}$

Thus $1+\lambda=3\mu$ and $3\lambda=1-2\mu$, giving $\lambda=1/11$, $\mu=4/11$ and the same OX. These parameters refer to the opposite lines from Method 1.

$\begin{aligned}&OY\\ & =\frac{2(3\mathbf a-\mathbf b)+11\mathbf b}{13}\\ & =\frac 3{13}(2\mathbf a+3\mathbf b)\\ & =\frac 3{13}OD.\end{aligned}$

Both vectors start at O and are parallel, so O,Y,D are collinear.

$BD=2\mathbf a+2\mathbf b$ and $BC=3\mathbf a-2\mathbf b$. Correctly expanding,

$\begin{aligned}BD\times BC&=6\mathbf a\times\mathbf a-4\mathbf a\times\mathbf b+6\mathbf b\times\mathbf a-4\mathbf b\times\mathbf b\\&=-10\mathbf a\times\mathbf b.\end{aligned}$

Hence $\operatorname{area}(BCD)=10\cdot\frac 12|\mathbf a\times\mathbf b|$, while $\operatorname{area}(OAB)=\frac 12|\mathbf a\times\mathbf b|$. The ratio is $1:10$.