If $$\hat{a}$$ is a unit vector such that $$(\bar{x}-\hat{a}) \cdot(\bar{x}+\hat{a})=8$$, then $$|\bar{x}|=$$

Let $$\vec{v}=2 \hat{i}+2 \hat{j}-\hat{k}$$ and $$\bar{w}=\hat{i}+3 \hat{k}$$. If $$\bar{u}$$ is a unit vector, then the maximum value of the scalar triple product $$[\bar{u} \bar{v} \bar{w}]$$ is

If $$\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \overline{\mathrm{b}}=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}, \overline{\mathrm{c}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}$$ and $$\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}$$ is perpendicular to $$\overline{\mathrm{c}}$$, then $$\lambda=$$

If $$3 \hat{j}, 4 \hat{k}$$ and $$3 \hat{j}+4 \hat{k}$$ are the position vectors of the vertices $$A, B, C$$ respectively of $$\triangle A B C$$, then the position vector of the point in which the bisector of $$\angle \mathrm{A}$$ meets $$\mathrm{BC}$$ is