$\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}, \hat{\mathbf{b}}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{c}}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$ are three vectors. If $\hat{\mathbf{d}}$ is a normal to the plane of $\hat{\mathbf{a}}$ and $\hat{\mathbf{b}}$ and d. $\hat{\mathbf{c}}=2$, then $|\hat{\mathbf{d}}|=$

If $\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \mathbf{c}=-\hat{\mathbf{k}}$ are position vectors of two points and $\mathbf{b}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}, \mathbf{d}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ are two vectors, then the lines $\mathbf{r}=\mathbf{a}+t \mathbf{b}, \mathbf{r}=\mathbf{c}+s \mathbf{d}$ are

$\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three vectors each having $\sqrt{2}$ magnitude such that $(\mathbf{a}, \mathbf{b})=(\mathbf{b}, \mathbf{c})=(\mathbf{c}, \mathbf{a})=\frac{\pi}{3}$. If $\mathbf{x}=\mathbf{a} \times(\mathbf{b} \times \mathbf{c})$ and $\mathbf{y}=\mathbf{b} \times(\mathbf{c} \times \mathbf{a})$, then