The vector equation of the line whose Cartesian equations are $$y=2$$ and $$4 x-3 z+5=0$$ is

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

If $${\pi \over 2} < \theta < \pi $$ and $$|\overline a | = 5,|\overline b | = 13,|\overline a \times \overline b | = 25$$, then the value of $$\overline a \,.\,\overline b $$ is

If $$|\bar{a} \times \bar{b}|^2+(\bar{a} \cdot \bar{b})^2=144$$ and $$|\bar{a}|=4$$, then $$|\bar{b}|=$$