Two lines $${L_1}:x = 5,{y \over {3 - \alpha }} = {z \over { - 2}}$$ and $${L_2}:x = \alpha ,{y \over { - 1}} = {z \over {2 - \alpha }}$$ are coplanar. Then $$\alpha $$ can take value(s)

A line $$l$$ passing through the origin is perpendicular to the lines $$$\,{l_1}:\left( {3 + t} \right)\widehat i + \left( { - 1 + 2t} \right)\widehat j + \left( {4 + 2t} \right)\widehat k,\,\,\,\,\, - \infty < t < \infty $$$ $$${l_2}:\left( {3 + 2s} \right)\widehat i + \left( {3 + 2s} \right)\widehat j + \left( {2 + s} \right)\widehat k,\,\,\,\,\, - \infty < s < \infty $$$
Then, the coordinate(s) of the points(s) on $${l_2}$$ at a distance of $$\sqrt {17} $$ from the point of intersection of $$l$$ and $${l_1}$$ is (are)

If the straight lines $$\,{{x - 1} \over 2} = {{y + 1} \over k} = {z \over 2}$$ and $${{x + 1} \over 5} = {{y + 1} \over 2} = {z \over k}$$ are coplanar, then the plane (s) containing these two lines is (are)

The vector (s) which is/are coplanar with vectors $${\widehat i + \widehat j + 2\widehat k}$$ and $${\widehat i + 2\widehat j + \widehat k,}$$ and perpendicular to the vector $${\widehat i + \widehat j + \widehat k}$$ is/are