Match the statements of Column I with values of Column II.

(A) Equation of the line passing through origin is \begin{aligned} & \frac{x}{a}=\frac{y}{b}=\frac{z}{c} \\ & \therefore \quad\left|\begin{array}{ccc} 2 & 1 & -1 \\ 1 & -2 & 1 \\ a & b & c \end{array}\right|=0 \\ & \Rightarrow \quad a(-1)-b(3)+c(-5)=0 \\ & \Rightarrow \quad-a-3 b-5 c=0 \\ & \Rightarrow \quad a+3 b+5 c=0 \\ & \text{ Also, } \quad\left|\begin{array}{ccc} \frac{8}{3} & -3 & 1 \\ 2 & -1 & 1 \\ a & b & c \end{array}\right|=0 \end{aligned}$$\begin{aligned} & \therefore \quad a(-2)-b\left(\frac{2}{3}\right)+c\left(\frac{10}{3}\right)=0 \\ & \Rightarrow \quad 2 a+\frac{2 b}{3}-\frac{10 c}{3}=0 \\ & 3 a+b-5 c=0 \\ & \end{aligned} $$From Eqs. (i) and (ii),$$ $$\begin{aligned} \frac{a}{-20} & =\frac{b}{20}=\frac{c}{-8} \\ \frac{a}{5} & =\frac{b}{-5}=\frac{c}{4} \end{aligned}$$ $$Equation of line is$$ \frac{x}{5}=\frac{y}{-5}=\frac{z}{4}=\lambda \text{ (say) } Also, $\frac{x-2}{1}=\frac{y-1}{2}=\frac{z+1}{1}=k_1$ (say) Now, $\frac{x-\frac{8}{3}}{2}=\frac{y+3}{-1}=\frac{z-1}{1}=k_2$ (say) Point on (iii) is $(5 \lambda,-5 \lambda,+4 \lambda)$ Point on (iv) is \left(2+k_1, 1-2 k_1,-1+k_1\right) Point on $(\mathrm{v})$ is \left(\frac{8}{3}+2 k_2,-3-k_2, 1+k_2\right) On solving, $2+k_1+1-2 k_1=0$ -k_1+3=0 $$\begin{aligned} k_1 & =3 \\ P & \equiv(5,-5,2) \end{aligned}$$ Again, for $Q$ $$\begin{array}{r} \frac{8}{3}+2 k_2-3-k_2=0 \\ k_2-\frac{1}{3}=0 \\ k_2=\frac{1}{3} \\ Q \equiv\left(\frac{10}{3}, \frac{-10}{3}, \frac{4}{3}\right) \end{array}$$ \text{ Now, } $$\begin{aligned} P Q & =\sqrt{\left(\frac{5}{3}\right)^2+\left(\frac{5}{3}\right)^2+\left(\frac{2}{3}\right)^2} \\ & =\frac{\sqrt{54}}{3} \end{aligned}$$ P Q^2=d^2=\frac{54}{9}=6 (B) $\tan ^{-1}\left(\frac{x+3-x+3}{1+\left(x^2-9\right)}\right)=\tan ^{-1}\left(\frac{3}{4}\right)$\begin{array}{rlrl} \Rightarrow & & \frac{6}{x^2-8} & =\frac{3}{4} \\ \Rightarrow & & 3 x^2 & =48 \\ \Rightarrow & x & =\pm 4 \end{array} (C) \begin{aligned} & (\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}}) \cdot\left(\overrightarrow{\mathbf{b}}+\frac{\overrightarrow{\mathbf{a}}-\mu \overrightarrow{\mathbf{b}}}{4}\right)=0 \\ & \Rightarrow \quad(\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}}) \cdot(4 \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{a}}-\mu \overrightarrow{\mathbf{b}})=0 \\ & \text{ Also, } 2\left|\overrightarrow{\mathbf{b}}+\frac{\overrightarrow{\mathbf{a}}-\mu \overrightarrow{\mathbf{b}}}{4}\right|=|\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}}| \\ & \Rightarrow \quad 2\left|\frac{(4-\mu) \overrightarrow{\mathbf{b}}^2-\overrightarrow{\mathbf{a}}^2=0}{4}\right|=|\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}}| \end{aligned}\begin{aligned} & \Rightarrow \frac{(4-\mu)^2 \overrightarrow{\mathbf{b}}^2}{4}+\frac{\overrightarrow{\mathbf{a}}^2}{4}=\overrightarrow{\mathbf{b}}^2+\overrightarrow{\mathbf{a}}^2 \\ & \Rightarrow \quad \frac{3 \overrightarrow{\mathbf{a}}^2}{4}=\frac{(4-\mu)^2-4}{4} \cdot \overrightarrow{\mathbf{a}}^2 \\ & 3 \overrightarrow{\mathbf{a}}^2=C(4-\mu)^2-4 \overrightarrow{\mathbf{b}}^2 \end{aligned} From Eqs. (i) and (ii) \begin{aligned} & 3(4-\mu)=(4-\mu)^2-4 \\ & (4-\mu)^2-3(4-\mu)-4=0 \\ & \Rightarrow \quad \mu=0,5 \\ & \mu=5 \text{ is not admissible. } \end{aligned} (D) $f(0)=9$, \begin{aligned} & f(x)=\frac{\sin \left(\frac{9 x}{2}\right)}{\sin \frac{x}{2}} \\ & =\left(3-4 \sin ^2 \frac{x}{2}\right)\left(3-4 \sin ^2 \frac{3 x}{2}\right) \end{aligned}\begin{aligned} & =9-12 \sin ^2 \frac{x}{2}-12 \sin ^2 \frac{3 x}{2} \\ & \quad+16 \sin ^2 \frac{x}{2} \cdot \sin ^2 \frac{3 x}{2} \\ & =9-6(1-\cos x)-6(1-\cos 3 x) \\ & \quad+4(1-\cos x)(1-\cos 3 x) \\ & =1+6 \cos x+6 \cos 3 x-4 \cos x \\ & \quad-4 \cos 3 x+4 \cos x \cos 3 x \\ & \text{ Let } \quad I=\frac{2}{\pi} \int_{-\pi}^\pi \frac{\sin \frac{9 x}{2}}{2} d x \\ & =\frac{4}{\pi} \int_0^\pi 1+2 \cos x+2 \cos 3 x \\ & =\frac{4}{\pi} \times \pi \quad+2(\cos 4 x+\cos x) d x \\ & =4 \end{aligned}$