$\mathbf{a} \cdot \mathbf{b}$ 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 $|\mathbf{r}|=$
The mean deviation from the median for the following data is
$$ \begin{array}{llllll} \hline x_1 & 9 & 3 & 7 & 2 & 5 \\ \hline f_1 & 1 & 6 & 2 & 8 & 4 \\ \hline \end{array} $$
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