Let $$X$$ and $$Y$$ be two events such that $$P\left( {X|Y} \right) = {1 \over 2},$$ $$P\left( {Y|X} \right) = {1 \over 3}$$ and $$P\left( {X \cap Y} \right) = {1 \over 6}.$$ Which of the following is (are) correct ?

If $$\overrightarrow a $$ and $$\overrightarrow b $$ are vectors such that $$\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt {29} $$ and $$\,\overrightarrow a \times \left( {2\widehat i + 3\widehat j + 4\widehat k} \right) = \left( {2\widehat i + 3\widehat j + 4\widehat k} \right) \times \widehat b,$$ then a possible value of $$\left( {\overrightarrow a + \overrightarrow b } \right).\left( { - 7\widehat i + 2\widehat j + 3\widehat k} \right)$$ is

The equation of a plane passing through the line of intersection of the planes $$x+2y+3z=2$$ and $$x-y+z=3$$ and at a distance $${2 \over {\sqrt 3 }}$$ from the point $$(3, 1, -1)$$ is

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)