30. Line $L_1$ passes through the point $\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and $\hat{\mathbf{k}}-\hat{\mathbf{i}}$. Line $L_2$ passes through the point $\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and is parallel to the vector $\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$. If $x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$ is the point of intersection of the lines $L_1$ and $L_2$, then $(y-x)=$

$\mathbf{a} \cdot \mathbf{b}$ 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 $|\mathbf{r}|=$