If $$\overline{\mathrm{a}}, \overline{\mathrm{b}} , \overline{\mathrm{c}}$$ are three vectors which are perpendicular to $$\overline{\mathrm{b}}+\overline{\mathrm{c}}, \overline{\mathrm{c}}+\overline{\mathrm{a}}$$ and $$\overline{\mathrm{a}}+\overline{\mathrm{b}}$$ respectively, such that $$|\bar{a}|=2,|\bar{b}|=3,|\bar{c}|=4$$, then $$|\bar{a}+\bar{b}+\bar{c}|=$$

$$(2 \hat{\mathrm{i}}+6 \hat{\mathrm{i}}+27 \hat{\mathrm{k}}) \times(\hat{\mathrm{i}}+\lambda \hat{\mathrm{j}}+\mu \hat{\mathrm{k}})=\overline{0}$$, then $$\lambda$$ and $$\mu$$ are respectively

If $\mathbf{a}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathbf{b}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+7 \hat{\mathbf{k}}$ and $\mathbf{c}=7 \hat{\mathbf{i}}-\hat{\mathbf{j}}+23 \hat{\mathbf{k}}$ are three vectors, then which of the following statement is true.

$\mathbf{a}$ and $\mathbf{b}$ are non-collinear vectors. If $p=(2 x+1) a-b$ and $q=(x-2) a+b$ are collinear vectors, then $x=$