The point of intersection of the lines represented by $\mathbf{r}=(\hat{\mathbf{i}}-6 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})+\mathbf{t}(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})$ and $\mathbf{r}=(4 \hat{\mathbf{j}}+\hat{\mathbf{k}})+\mathbf{s}(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}})$ is

$\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three vectors such that $|\mathbf{a}|=2,|\mathbf{b}|=3$, $|\mathbf{c}|=5,|\mathbf{a}+\mathbf{b}+\mathbf{c}|=\sqrt{69}$. If $(\mathbf{a} \cdot \mathbf{b})=(\mathbf{b} \cdot \mathbf{c})=\frac{\pi}{3}$, then $(\mathbf{c}, \mathbf{a})=$

If the points $A, B, C, D$ with positions vectors $\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}, \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, 2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ respectively form a tetrahedron, then the angle between the faces $A B C$ and $A B D$ of the tetrahedron is

$\mathbf{a}, \mathbf{b}, \mathbf{c}$ are unit vectors. If $\mathbf{a}, \mathbf{b}$ are perpendicular vectors, $(\mathbf{a}-\mathbf{c}) \cdot(\mathbf{b}+\mathbf{c})=0$ and $\mathbf{c}=l \mathbf{a}+m \mathbf{b}+n(\mathbf{a} \times \mathbf{b}) ;$ ( $l, m, n$ are scalars), then $n^2=$