If $\mathbf{b}, \mathbf{c}$ are non collinear vectors, $|\mathbf{c}| \neq 0$, $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})+(\mathbf{a} \cdot \mathbf{b}) \mathbf{b}=(4-2 \beta-\sin \alpha) \mathbf{b}+\left(\beta^2-1\right) \mathbf{c}$ and $(\mathbf{c} \cdot \mathbf{c}) \mathbf{a}=\mathbf{c}$, then the scalars $\alpha$ and $\beta$ are

If $12 \hat{\mathbf{i}}-12 \hat{\mathbf{j}}-18 \hat{\mathbf{k}},-3 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}-9 \hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-24 \hat{\mathbf{k}}$ be the position vectors of the vertices $A, B$ and $C$ respectively of $\triangle A B C$, then the position vector of the incentre of $\triangle A B C$ is

For non-coplanar vectors $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$, if the point of intersection of the line $\mathbf{r}=\mathbf{a}+t(\mathbf{b}-\mathbf{c})$ and the plane $\mathbf{r}=\mathbf{b}+\mathbf{c}+x(\mathbf{a}-\mathbf{b})+y(\mathbf{c}+\mathbf{a})$ is $l \mathbf{a}+m \mathbf{b}+n \mathbf{c}$, then $3 l+4 m+2 n=$

If the orthocentre of the triangle whose vertices are $2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}, 5 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ is $x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$, then