Let $(x, y) \in R \times R$ and $\mathbf{a}=x \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathbf{b}=6 \hat{\mathbf{i}}-y\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ be two vectors. If

$$ |\mathbf{a} \times \mathbf{b}|^2+|\mathbf{a} \cdot \mathbf{b}|^2=f(x) g(y), \text { then } f(x)+g(y)-46=0 $$

represents

$\mathbf{a} \cdot \mathbf{b}$ and $\mathbf{c}$ are the position vectors of three non-collinear points on a plane. If

$$ \alpha=[\mathbf{a b c}] \text { and } \mathbf{r}=\mathbf{a} \times \mathbf{b}-\mathbf{c} \times \mathbf{b}-\mathbf{a} \times \mathbf{c} \text {, then }\left|\frac{\alpha}{\mathbf{r}}\right| $$

represents

If $P=(\mathbf{a} \times \hat{\mathbf{i}})^2+(\mathbf{a} \times \hat{\mathbf{j}})^2+(\mathbf{a} \times \hat{\mathbf{k}})^2$ and $Q=(\mathbf{a} \cdot \hat{\mathbf{i}})^2+(\mathbf{a} \cdot \hat{\mathbf{j}})^2+(\mathbf{a} \cdot \hat{\mathbf{k}})^2$, then

$\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}, \mathbf{c}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ are three vectors. If $\mathbf{r}$ is a vector such that $\mathbf{r} \cdot \mathbf{a}=0, \mathbf{r} \cdot \mathbf{c}=3$ and $\left[\begin{array}{ll}\mathbf{r} & \mathbf{a} \\ \mathbf{b}\end{array}\right]=0$, then $|\mathbf{r}|=$