If $\mathbf{u}, \mathbf{v}, \mathbf{w}$ are non-coplanar vectors and $p, q$ are real numbers, then the equality $[3 \mathbf{u} p \mathbf{v} p \mathbf{w}]-[p \mathbf{v} \mathbf{w} q \mathbf{u}]-[2 \mathbf{w} q \mathbf{v} q \mathbf{u}]=0$ holds for

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