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)

Let $${\overrightarrow A }$$ be vector parallel to line of intersection of planes $${P_1}$$ and $${P_2}.$$ Planes $${P_1}$$ is parallel to the vectors $$2\widehat j + 3\widehat k$$ and $$4\widehat j - 3\widehat k$$ and that $${P_2}$$ is parallel to $$\widehat j - \widehat k$$ and $$3\widehat i + 3\widehat j,$$ then the angle between vector $${\overrightarrow A }$$ and a given vector $$2\widehat i + \widehat j - 2\widehat k$$ is