$$(1 + 2\lambda ) + 2 - 3\lambda + 3 - 6\lambda = 5$$

$$ \Rightarrow 6 - 7\lambda = 5 \Rightarrow \lambda = {1 \over 7}$$

so, $$P = \left( {{9 \over 7}, - {{11} \over 7},{{15} \over 7}} \right)$$

$$AP = \sqrt {{{\left( {1 - {9 \over 7}} \right)}^2} + {{\left( { - 2 + {{11} \over 7}} \right)}^2} + {{\left( {3 - {{15} \over 7}} \right)}^2}} $$

$$AP = \sqrt {\left( {{4 \over {49}}} \right) + {9 \over {49}} + {{36} \over {49}}} = 1$$

$$2\sin \left( {{\pi \over 8}} \right)\sin \left( {{{2\pi } \over 8}} \right)\sin \left( {{{3\pi } \over 8}} \right)\sin \left( {{{5\pi } \over 8}} \right)\sin \left( {{{6\pi } \over 8}} \right)\sin \left( {{{7\pi } \over 8}} \right)$$

$$2{\sin ^2}{\pi \over 8}{\sin ^2}{{2\pi } \over 8}{\sin ^2}{{3\pi } \over 8}$$

$${\sin ^2}{\pi \over 8}{\sin ^2}{{3\pi } \over 8}$$

$${\sin ^2}{\pi \over 8}{\cos ^2}{\pi \over 8}$$

$${1 \over 4}{\sin ^2}\left( {{\pi \over 4}} \right) = {1 \over 8}$$

$${{\cos x} \over {1 + \sin x}} = \left| {\tan 2x} \right|$$

$$ \Rightarrow {{{{\cos }^2}x/2 - {{\sin }^2}x/2} \over {(\cos x/2 + \sin x/2)^2}} = \left| {\tan 2x} \right|$$

$$ \Rightarrow $$ $${{\cos {x \over 2} - \sin {x \over 2}} \over {\cos {x \over 2} + \sin {x \over 2}}} = \left| {\tan 2x} \right|$$

$$ \Rightarrow $$ $${{1 - \tan {x \over 2}} \over {1 + \tan {x \over 2}}} = \left| {\tan 2x} \right|$$

$$ \Rightarrow $$ $${{\tan {\pi \over 4} - \tan {x \over 2}} \over {\tan {\pi \over 4} + \tan {x \over 2}}} = \left| {\tan 2x} \right|$$

$$ \Rightarrow {\tan ^2}\left( {{\pi \over 4} - {\pi \over 2}} \right) = {\tan ^2}2x$$

$$ \Rightarrow 2x = n\pi \pm \left( {{\pi \over 4} - {\pi \over 2}} \right)$$

$$ \Rightarrow x = {{ - 3\pi } \over {10}},{{ - \pi } \over 6},{\pi \over {10}}$$

or sum $$ = {{ - 11\pi } \over 6}$$.

$$\sin \theta + \cos \theta = {1 \over 2}$$

$${\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta = {1 \over 4}$$

$$\sin 2\theta = - {3 \over 4}$$

Now :

$$\cos 4\theta = 1 - 2{\sin ^2}2\theta $$

$$ = 1 - 2{\left( { - {3 \over 4}} \right)^2}$$

$$ = 1 - 2 \times {9 \over {16}} = - {1 \over 8}$$

$$\sin 6\theta = 3\sin 2\theta - 4{\sin ^3}2\theta $$

$$ = (3 - 4{\sin ^2}2\theta ).\sin 2\theta $$

$$ = \left[ {3 - 4\left( {{9 \over {16}}} \right)} \right].\left( { - {3 \over 4}} \right)$$

$$ \Rightarrow \left[ {{3 \over 4}} \right] \times \left( { - {3 \over 4}} \right) = - {9 \over {16}}$$

$$16[\sin 2\theta + \cos 4\theta + \sin 6\theta ]$$

= $$16\left( { - {3 \over 4} - {1 \over 8} - {9 \over {16}}} \right) = - 23$$