Let the line $$\,\,\,\,\,$$ $${{x - 2} \over 3} = {{y - 1} \over { - 5}} = {{z + 2} \over 2}$$ lie in the plane $$\,\,\,\,\,$$ $$x + 3y - \alpha z + \beta = 0.$$ Then $$\left( {\alpha ,\beta } \right)$$ equals

The projections of a vector on the three coordinate axis are $$6,-3,2$$ respectively. The direction cosines of the vector are :

The line passing through the points $$(5,1,a)$$ and $$(3, b, 1)$$ crosses the $$yz$$-plane at the point $$\left( {0,{{17} \over 2}, - {{ - 13} \over 2}} \right)$$ . Then

If the straight lines $$\,\,\,\,\,$$ $$\,\,\,\,\,$$ $${{x - 1} \over k} = {{y - 2} \over 2} = {{z - 3} \over 3}$$ $$\,\,\,\,\,$$ and$$\,\,\,\,\,$$ $${{x - 2} \over 3} = {{y - 3} \over k} = {{z - 1} \over 2}$$ intersects at a point, then the integer $$k$$ is equal to