If $\overline{\mathrm{a}}=\hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\overline{\mathrm{c}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}$, then the vector $\overline{\mathrm{b}}$ satisfying $\overline{\mathrm{a}} \times \overline{\mathrm{b}}+\overline{\mathrm{c}}=\overline{0}$ and $\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}=3$ is

$\overline{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overline{\mathrm{b}}=4 \hat{\mathrm{i}}-2 \hat{j}+3 \hat{k}, \overline{\mathrm{c}}=\hat{i}-2 \hat{j}+\hat{k}$, then $a$ vector of magnitude 6 units, which is parallel to the vector $2 \bar{a}-\bar{b}+3 c$, is

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

If $\quad \overline{\mathrm{a}}=\hat{\mathrm{i}}-\hat{\mathrm{k}}, \overline{\mathrm{b}}=x \hat{\mathrm{i}}+\hat{\mathrm{j}}+(1-x) \hat{\mathrm{k}} \quad$ and $\overline{\mathrm{c}}=y \hat{\mathrm{i}}+x \hat{\mathrm{j}}+(1+x-y) \hat{\mathrm{k}}$ then $\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})$ depends on