(i) Find the equation of the plane passing through the points $$(2, 1, 0), (5, 0, 1)$$ and $$(4, 1, 1).$$
(ii) If $$P$$ is the point $$(2, 1, 6)$$ then find the point $$Q$$ such that $$PQ$$ is perpendicular to the plane in (i) and the mid point of $$PQ$$ lies on it.

If $$\overrightarrow u ,\overrightarrow v ,\overrightarrow w ,$$ are three non-coplanar unit vectors and $$\alpha ,\beta ,\gamma $$ are the angles between $$\overrightarrow u $$ and $$\overrightarrow v $$ and $$\overrightarrow w ,$$ $$\overrightarrow w $$ and $$\overrightarrow u $$ respectively and $$\overrightarrow x ,\overrightarrow y ,\overrightarrow z ,$$ are unit vectors along the bisectors of the angles $$\alpha ,\,\,\beta ,\,\,\gamma $$ respectively. Prove that $$\,\left[ {\overrightarrow x \times \overrightarrow y \,\,\overrightarrow y \times \overrightarrow z \,\,\overrightarrow z \times \overrightarrow x } \right] = {1 \over {16}}{\left[ {\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow w } \right]^2}\,{\sec ^2}{\alpha \over 2}{\sec ^2}{\beta \over 2}{\sec ^2}{\gamma \over 2}.$$

For a student to qualify, he must pass at least two out of three exams. The probability that he will pass the 1^st exam is $$p.$$ If he fails in one of the exams then the probability of his passing in the next exam is $${p \over 2}$$ otherwise it remains the same. Find the probability that he will qualify.

$$A$$ is targeting to $$B, B$$ and $$C$$ are targeting to $$A.$$ Probability of hitting the target by $$A,B$$ and $$C$$ are $${2 \over 3},{1 \over 2}$$ and $${1 \over 3}$$ respectively. If $$A$$ is hit then find the probability that $$B$$ hits the target and $$C$$ does not.