The polar co-ordinates of the point, whose Cartesian coordinates are $$(-2 \sqrt{3}, 2)$$, are

If the vectors $$2 \hat{i}-\hat{j}-\hat{k} ; \hat{i}+2 \hat{j}-3 \hat{k}$$ and $$3 \hat{i}+\lambda \hat{j}+5 \hat{k}$$ are coplanar, then the value of $$\lambda$$ is

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 $$\lambda=$$