(A) 

 $\frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-1}{-2}=\frac{\mathrm{z}+1}{1}=\mathrm{r}$

 $\therefore$ P (r + 2– 2 r + 1, r –1)

 $\frac{\mathrm{x}}{\mathrm{r}+2}=\frac{\mathrm{y}}{1-2\mathrm{r}}=\frac{\mathrm{z}}{\mathrm{r}-1}=\mathrm{K}$

 $\frac{\mathrm{K}(\mathrm{r}+2)-8/3}{2}=\frac{\mathrm{y}}{1-2\mathrm{r}}=\frac{\mathrm{z}}{\mathrm{r}-1}=\mathrm{K}$

K (r –1 + 1 – 2r ) = –2

K (–r) = –2

Kr = 2

 $\frac{2+2\mathrm{K}=8/3}{2}=\frac{\mathrm{K}-2.2+3}{-1}$

 $\frac{2\mathrm{K}-2/3}{2}=\frac{\mathrm{K}-1}{-1}$

 $\frac{1}{3}+\mathrm{K}-1=0$

 $\therefore \mathrm{K}=\frac{2}{3}$

 $\therefore \mathrm{r} = 3$

 $\mathrm{P}(5,-5,2)\mathrm{Q}(5.\frac{2}{3},-5.\frac{2}{3},2.\frac{2}{3})$

 $\mathrm{d}=\sqrt{5^2+5^2+2^2}.\frac{1}{3}$

 $\Rightarrow {\mathrm{d}}^2=\frac{54}{9}=6$

(B)

 ${\tan }^{-1}(\mathrm{x}+3)-{\tan }^{-1}(\mathrm{x}-3)={\sin }^{-1}\left(\frac{3}{5}\right)$

 $\frac{6}{1+{\mathrm{x}}^2-9}=\frac{3}{4}$

8 = x² – 8

 $\therefore$x = ± 4

(C)

 $\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=0$

 $\overrightarrow{\mathrm{a}}=\mathrm{\mu }\overrightarrow{\mathrm{b}}+\mathrm{\mu }\overrightarrow{\mathrm{c}}$

 $\overrightarrow{\mathrm{c}}=\frac{\overrightarrow{\mathrm{a}}-\mathrm{\mu }\overrightarrow{\mathrm{b}}}{4}$

 $\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\mathrm{b}+\frac{\overrightarrow{\mathrm{a}}-\mathrm{\mu }\overrightarrow{\mathrm{b}}}{4}$

 $\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\frac{(4-\mathrm{\mu })\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{a}}}{4}$

 $(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}).(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=0$

 $\text{(}\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}})\left(\frac{(4-\mathrm{\mu })\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{a}}}{4}\right)=0$

 $(4-\mathrm{\mu })|\overrightarrow{\mathrm{b}}{|}^2=|\overrightarrow{\mathrm{a}}{|}^2$

so µ < 4

 $\left|\frac{(4-\mathrm{\mu })\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{a}}}{4}\right|=|\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}|$

(4 – µ)² b² + a² = 4 (b² + a²)

3a² = ((4–µ)² – 4) b²

3(4–µ) = (16 + µ² – 8 µ – 4)

12 – 3 µ = µ¹ – 8µ+12

µ² – 5µ = 0

µ (µ–5) = 0 $\Rightarrow$ µ = 0

µ $\ne$ 5 as µ < 4

(D)

 $\mathrm{I}=\frac{2}{\mathrm{\pi }}\underset{-\mathrm{\pi }}{\overset{\mathrm{\pi }}{\int }}\frac{\sin \left(\frac{9\mathrm{x}}{2}\right)}{\sin \left(\frac{\mathrm{x}}{2}\right)}\mathrm{dx}$

 $\mathrm{I}=\frac{4}{\mathrm{\pi }}\underset{-\mathrm{\pi }}{\overset{\mathrm{\pi }}{\int }}\frac{\sin \left(\frac{9\mathrm{x}}{2}\right)}{\sin \left(\frac{\mathrm{x}}{2}\right)}\mathrm{dx}$

 $\mathrm{I}=\frac{4}{\mathrm{\pi }}\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{\cos \left(\frac{9\mathrm{x}}{2}\right)}{\cos \left(\frac{\mathrm{x}}{2}\right)}\mathrm{dx}$

 $2\mathrm{I}=\frac{4}{\mathrm{\pi }}\underset{0}{\overset{\mathrm{\pi }}{\int }}(\frac{\sin \frac{9\mathrm{x}}{2}}{\sin \frac{\mathrm{x}}{2}}+\frac{\cos \frac{9\mathrm{x}}{2}}{\cos \frac{\mathrm{x}}{2}})\mathrm{dx}=\frac{4}{\mathrm{\pi }}\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{\sin (\frac{9\mathrm{x}}{2}+\frac{\mathrm{x}}{2})}{2\sin \frac{\mathrm{x}}{2}\cos \frac{\mathrm{x}}{2}}\mathrm{dx}=\frac{8}{\mathrm{\pi }}\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{\sin 5\mathrm{x}}{\mathrm{sinx}}\mathrm{dx}$

 $=\frac{8}{\mathrm{\pi }}\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{\sin 2\mathrm{xcos}3\mathrm{x}+\cos 2\mathrm{xsin}3\mathrm{x}}{\mathrm{sinx}}\mathrm{dx}=\frac{8}{\mathrm{\pi }}\underset{0}{\overset{\mathrm{\pi }}{\int }}2\mathrm{cosxcos}3\mathrm{x}+(3-4{\sin }^2\mathrm{x})\cos 2\mathrm{x}\mathrm{dx}$

 $=\frac{8}{\mathrm{\pi }}\underset{0}{\overset{\mathrm{\pi }}{\int }}(\cos 4\mathrm{x}+3\cos 2\mathrm{x}+\cos 3\mathrm{x}-\mathrm{cosx})\mathrm{dx}=\frac{8}{\mathrm{\pi }}\underset{0}{\overset{\mathrm{\pi }}{\int }}{[\frac{\sin 4\mathrm{x}}{4}+\frac{3\sin 2\mathrm{x}}{2}+\frac{\sin 3\mathrm{x}}{3}-\mathrm{sinx}]}_0^{\mathrm{\pi }}$