Let z₁ = x₁ + iy ; z₂ = x₂ + iy₂

z₁ $$-$$ z₂ = (x₁ $$-$$ x₂) + i(y₁ $$-$$ y₂)

$$\therefore$$ $$\arg ({z_1} - {z_2}) = {\pi \over 4}$$ $$\Rightarrow$$ $${\tan ^{ - 1}}\left( {{{{y_1} - {y_2}} \over {{x_1} - {x_2}}}} \right) = {\pi \over 4}$$

$${y_1} - {y_2} = {x_1} - {x_2}$$ ....... (1)

$$|{z_1} - 3|\, = {\mathop{\rm Re}\nolimits} ({z_1}) \Rightarrow {({x_1} - 3)^2} + {y_1}^2 = {x_1}^2$$ .... (2)

$$|{z_2} - 3|\, = {\mathop{\rm Re}\nolimits} ({z_2}) \Rightarrow {({x_2} - 3)^2} + {y_2}^2 = {x_2}^2$$ .... (3)

sub (2) & (3)

$${({x_1} - 3)^2} - {({x_2} - 3)^2} + {y_1}^2 - {y_2}^2 = {x_1}^2 - {x_2}^2$$

$$({x_1} - {x_2})({x_1} + {x_2} - 6) + ({y_1} - {y_2})({y_1} + {y_2})$$

$$ = ({x_1} - {x_2})({x_1} + {x_2})$$

$${x_1} + {x_2} - 6 + {y_1} + {y_2} = {x_1} + {x_2} \Rightarrow {y_1} + {y_2} = 6$$

$$\sum {P(X) = 1 \Rightarrow k + 2k + 3} k + k = 1$$

$$ \Rightarrow k = {1 \over 9}$$

Now, $$p = P\left( {{{kx < 4} \over {X < 3}}} \right) = {{P(X = 2)} \over {P(X < 3)}} = {{{{2k} \over {9k}}} \over {{k \over {9k}} + {{2k} \over {9k}}}} = {2 \over 3}$$

$$ \Rightarrow p = {2 \over 3}$$

Now, $$5p = \lambda k$$

$$ \Rightarrow (5)\left( {{2 \over 3}} \right) = \lambda (1/9)$$

$$ \Rightarrow \lambda = 30$$

Given equation

$${\sin ^4}\theta + {\cos ^4}\theta - \sin \theta \cos \theta = 0$$

$$ \Rightarrow 1 - {\sin ^2}\theta {\cos ^2}\theta - \sin \theta \cos \theta = 0$$

$$ \Rightarrow 2 - {(\sin 2\theta )^2} - \sin 2\theta = 0$$

$$ \Rightarrow {(\sin 2\theta )^2} + (\sin 2\theta ) - 2 = 0$$

$$ \Rightarrow (\sin 2\theta + 2)(\sin 2\theta - 1) = 0$$

$$ \Rightarrow \sin 2\theta = 1$$ or $$\sin 2\theta = - 2$$ (Not Possible)

$$ \Rightarrow 2\theta = {\pi \over 2},{{5\pi } \over 2},{{9\pi } \over 2},{{13\pi } \over 2}$$

$$ \Rightarrow \theta = {\pi \over 4},{{5\pi } \over 4},{{9\pi } \over 4},{{13\pi } \over 4}$$

$$ \Rightarrow S = {\pi \over 4} + {{5\pi } \over 4} + {{9\pi } \over 4} + {{13\pi } \over 4} = 7\pi $$

$$ \Rightarrow {{8S} \over \pi } = {{8 \times 7\pi } \over \pi } = 56.00$$

Case I : When cot x > 0, $$x \in \left[ {0,{\pi \over 2}} \right] \cup \left[ {\pi ,{{3\pi } \over 2}} \right]$$

$$\cot x = \cot x + {1 \over {\sin x}} \Rightarrow $$ not possible

Case II : When cot x < 0, $$x \in \left[ {{\pi \over 2},\pi } \right] \cup \left[ {{{3\pi } \over 2},2\pi } \right]$$

$$ - \cot x = \cot x + {1 \over {\sin x}}$$

$$ \Rightarrow {{ - 2\cos x} \over {\sin x}} = {1 \over {\sin x}}$$

$$ \Rightarrow \cos x = {{ - 1} \over 2}$$

$$ \Rightarrow x = {{2\pi } \over 3},{{4\pi } \over 3}$$(Rejected)

One solution.