Part (A)

Let the two given lines be $L_1$ and $L_2$. $L_1: \frac{x-2}{1} = \frac{y-1}{-2} = \frac{z+1}{1} = \lambda$ $L_2: \frac{x - 8/3}{2} = \frac{y+3}{-1} = \frac{z-1}{1} = \gamma$

A general point on $L_1$ is $P(\lambda+2, -2\lambda+1, \lambda-1)$. A general point on $L_2$ is $Q(2\gamma+8/3, -\gamma-3, \gamma+1)$.

A line from the origin $O(0,0,0)$ meets $L_1$ at $P$ and $L_2$ at $Q$. This means the points $O$, $P$, and $Q$ are collinear. Therefore, their position vectors $\vec{OP}$ and $\vec{OQ}$ are parallel. $\vec{OP} = (\lambda+2)\hat{i} + (-2\lambda+1)\hat{j} + (\lambda-1)\hat{k}$ $\vec{OQ} = (2\gamma+8/3)\hat{i} + (-\gamma-3)\hat{j} + (\gamma+1)\hat{k}$

For collinearity, the direction ratios must be proportional: $\frac{\lambda+2}{2\gamma+8/3} = \frac{-2\lambda+1}{-\gamma-3} = \frac{\lambda-1}{\gamma+1}$

From the last two ratios: $(\-2\lambda+1)(\gamma+1) = (\lambda-1)(-\gamma-3)$ $-2\lambda\gamma - 2\lambda + \gamma + 1 = -\lambda\gamma - 3\lambda + \gamma + 3$ $-\lambda\gamma + \lambda - 2 = 0 \implies \lambda\gamma - \lambda + 2 = 0$ (1)

From the first and third ratios: $(\lambda+2)(\gamma+1) = (\lambda-1)(2\gamma+8/3)$ $\lambda\gamma + \lambda + 2\gamma + 2 = 2\lambda\gamma + \frac{8}{3}\lambda - 2\gamma - \frac{8}{3}$ $\lambda\gamma + \frac{5}{3}\lambda - 4\gamma - \frac{14}{3} = 0$ (2)

From (1), $\lambda\gamma = \lambda - 2$. Substituting this into (2): $(\lambda-2) + \frac{5}{3}\lambda - 4\gamma - \frac{14}{3} = 0$ $\frac{8}{3}\lambda - 4\gamma - \frac{20}{3} = 0 \implies 8\lambda - 12\gamma - 20 = 0 \implies 2\lambda - 3\gamma - 5 = 0$ (3)

From (1), if $\lambda \neq 0$, $\gamma = \frac{\lambda-2}{\lambda}$. Substituting into (3): $2\lambda - 3\left(\frac{\lambda-2}{\lambda}\right) - 5 = 0$ $2\lambda^2 - 3(\lambda-2) - 5\lambda = 0$ $2\lambda^2 - 3\lambda + 6 - 5\lambda = 0$ $2\lambda^2 - 8\lambda + 6 = 0$ $\lambda^2 - 4\lambda + 3 = 0$ $(\lambda-1)(\lambda-3) = 0 \implies \lambda=1$ or $\lambda=3$.

If $\lambda=1$, $\gamma = \frac{1-2}{1} = -1$. $P(3, -1, 0)$, $Q(2/3, -2, 0)$. Direction ratios are $(3,-1,0)$ and $(2/3,-2,0)$, which are not proportional. So $\lambda=1$ is not a solution. If $\lambda=3$, $\gamma = \frac{3-2}{3} = 1/3$. $P(5, -5, 2)$, $Q(10/3, -10/3, 4/3)$. Direction ratios are $(5,-5,2)$ and $(10/3, -10/3, 4/3)$, which are proportional ($10/3 = (2/3) \times 5$, etc.). This is the correct solution.

The points are $P(5, -5, 2)$ and $Q(10/3, -10/3, 4/3)$. The distance $PQ = d$ is given by: $d^2 = (5 - \frac{10}{3})^2 + (-5 - (-\frac{10}{3}))^2 + (2 - \frac{4}{3})^2$ $d^2 = (\frac{5}{3})^2 + (-\frac{5}{3})^2 + (\frac{2}{3})^2 = \frac{25}{9} + \frac{25}{9} + \frac{4}{9} = \frac{54}{9} = 6$. So, $d^2=6$. This matches (t).

Part (B)

The equation is ${\tan ^{ - 1}}\left( {x + 3} \right) - {\tan ^{ - 1}}\left( {x - 3} \right) = {\sin ^{ - 1}}\left( {{3 \over 5}} \right)$. Using the formula ${\tan ^{ - 1}}A - {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\frac{{A - B}}{{1 + AB}}} \right)$, which is valid if $AB > -1$. Let $A=x+3$ and $B=x-3$. $AB = (x+3)(x-3) = x^2-9$. The formula is valid if $x^2-9 > -1 \implies x^2 > 8$.

LHS = ${\tan ^{ - 1}}\left( {\frac{{(x + 3) - (x - 3)}}{{1 + (x + 3)(x - 3)}}} \right) = {\tan ^{ - 1}}\left( {\frac{6}{{1 + x^2 - 9}}} \right) = {\tan ^{ - 1}}\left( {\frac{6}{{x^2 - 8}}} \right)$. For the RHS, let $\theta = {\sin ^{ - 1}}(3/5)$. Then $\sin \theta = 3/5$. This corresponds to a right triangle with sides 3, 4, 5. So, $\tan \theta = 3/4$. Thus, ${\sin ^{ - 1}}(3/5) = {\tan ^{ - 1}}(3/4)$.

The equation becomes ${\tan ^{ - 1}}\left( {\frac{6}{{x^2 - 8}}} \right) = {\tan ^{ - 1}}\left( {{3 \over 4}} \right)$. $\frac{6}{{x^2 - 8}} = \frac{3}{4}$ $24 = 3(x^2 - 8) \implies 8 = x^2 - 8 \implies x^2 = 16 \implies x = \pm 4$.

We must check our initial condition $x^2 > 8$. For both $x=4$ and $x=-4$, $x^2=16$, which is greater than 8. So both solutions are valid. The values of $x$ are $4$ and $-4$. These match (r) and (p).

Part (C)

Given:

1. $\overrightarrow a \cdot \overrightarrow b = 0$
2. $(\overrightarrow b - \overrightarrow a) \cdot (\overrightarrow b + \overrightarrow c) = 0 \implies |\vec{b}|^2 + \vec{b}\cdot\vec{c} - \vec{a}\cdot\vec{b} - \vec{a}\cdot\vec{c} = 0 \implies |\vec{b}|^2 + \vec{b}\cdot\vec{c} - \vec{a}\cdot\vec{c} = 0$
3. $2|\overrightarrow b + \overrightarrow c| = |\overrightarrow b - \overrightarrow a| \implies 4|\overrightarrow b + \overrightarrow c|^2 = |\overrightarrow b - \overrightarrow a|^2 \implies 4(|\vec{b}|^2+2\vec{b}\cdot\vec{c}+|\vec{c}|^2) = |\vec{b}|^2 - 2\vec{a}\cdot\vec{b} + |\vec{a}|^2$. Using (1), this simplifies to $3|\vec{b}|^2 + 8\vec{b}\cdot\vec{c} + 4|\vec{c}|^2 = |\vec{a}|^2$.
4. $\overrightarrow a = \mu \overrightarrow b + 4\overrightarrow c$

From (1) and (4): $(\mu \overrightarrow b + 4\overrightarrow c) \cdot \overrightarrow b = 0 \implies \mu|\vec{b}|^2 + 4\vec{c}\cdot\vec{b} = 0 \implies \vec{b}\cdot\vec{c} = -\frac{\mu}{4}|\vec{b}|^2$. (I)

From (4), $|\vec{a}|^2 = (\mu\vec{b}+4\vec{c})\cdot(\mu\vec{b}+4\vec{c}) = \mu^2|\vec{b}|^2+8\mu\vec{b}\cdot\vec{c}+16|\vec{c}|^2$. (II)

Substitute (II) into the result from (3): $3|\vec{b}|^2 + 8\vec{b}\cdot\vec{c} + 4|\vec{c}|^2 = \mu^2|\vec{b}|^2+8\mu\vec{b}\cdot\vec{c}+16|\vec{c}|^2$ $(3-\mu^2)|\vec{b}|^2 + (8-8\mu)\vec{b}\cdot\vec{c} - 12|\vec{c}|^2 = 0$. Substitute (I) into this equation: $(3-\mu^2)|\vec{b}|^2 + 8(1-\mu)(-\frac{\mu}{4}|\vec{b}|^2) - 12|\vec{c}|^2 = 0$ $(3-\mu^2 - 2\mu(1-\mu))|\vec{b}|^2 = 12|\vec{c}|^2$ $(\mu^2 - 2\mu + 3)|\vec{b}|^2 = 12|\vec{c}|^2$. (III)

Now use relation from (2): $|\vec{b}|^2 + \vec{b}\cdot\vec{c} - (\mu\vec{b}+4\vec{c})\cdot\vec{c} = 0$ $|\vec{b}|^2 + \vec{b}\cdot\vec{c} - (\mu\vec{b}\cdot\vec{c}+4|\vec{c}|^2) = 0$ $|\vec{b}|^2 + (1-\mu)\vec{b}\cdot\vec{c} - 4|\vec{c}|^2 = 0$. Substitute (I): $|\vec{b}|^2 + (1-\mu)(-\frac{\mu}{4}|\vec{b}|^2) - 4|\vec{c}|^2 = 0$ $(1 - \frac{\mu-\mu^2}{4})|\vec{b}|^2 = 4|\vec{c}|^2$ $(\mu^2 - \mu + 4)|\vec{b}|^2 = 16|\vec{c}|^2$. (IV)

From (III) and (IV), since $\vec{b}, \vec{c}$ are non-zero: $\frac{|\vec{c}|^2}{|\vec{b}|^2} = \frac{\mu^2 - 2\mu + 3}{12} = \frac{\mu^2 - \mu + 4}{16}$ $16(\mu^2 - 2\mu + 3) = 12(\mu^2 - \mu + 4)$ $4(\mu^2 - 2\mu + 3) = 3(\mu^2 - \mu + 4)$ $4\mu^2 - 8\mu + 12 = 3\mu^2 - 3\mu + 12$ $\mu^2 - 5\mu = 0 \implies \mu(\mu-5)=0$. The possible values for $\mu$ are $0$ and $5$. These match (q) and (s).

Part (D)

We need to evaluate $V = \frac{2}{\pi} \int_{-\pi}^{\pi} f(x) dx$, where $f(x) = \frac{\sin(9x/2)}{\sin(x/2)}$ for $x \ne 0$ and $f(0)=9$. The function $f(x)$ is an even function since $f(-x) = \frac{\sin(-9x/2)}{\sin(-x/2)} = \frac{-\sin(9x/2)}{-\sin(x/2)} = f(x)$. So, $\int_{-\pi}^{\pi} f(x) dx = 2 \int_{0}^{\pi} f(x) dx$.

We use the identity for the Dirichlet kernel: $1 + 2\sum_{k=1}^n \cos(kx) = \frac{\sin((n+1/2)x)}{\sin(x/2)}$. For our function, we have $n+1/2 = 9/2$, which gives $n=4$. So, $f(x) = 1 + 2\sum_{k=1}^4 \cos(kx) = 1 + 2(\cos x + \cos 2x + \cos 3x + \cos 4x)$.

Now we integrate: $\int_{-\pi}^{\pi} f(x) dx = \int_{-\pi}^{\pi} (1 + 2\sum_{k=1}^4 \cos(kx)) dx$ $= \int_{-\pi}^{\pi} 1 dx + 2\sum_{k=1}^4 \int_{-\pi}^{\pi} \cos(kx) dx$. For any integer $k \ge 1$, $\int_{-\pi}^{\pi} \cos(kx) dx = [\frac{\sin(kx)}{k}]_{-\pi}^{\pi} = 0$. So, $\int_{-\pi}^{\pi} f(x) dx = \int_{-\pi}^{\pi} 1 dx = [x]_{-\pi}^{\pi} = \pi - (-\pi) = 2\pi$.

The required value is $V = \frac{2}{\pi} (2\pi) = 4$. This matches (r).

Summary of Matches

• (A) $d^2=6 \implies$ (t)
• (B) $x = -4, 4 \implies$ (p), (r)
• (C) $\mu = 0, 5 \implies$ (q), (s)
• (D) Value is $4 \implies$ (r)

This combination corresponds to option A: $(A) \to t; (B) \to p,r; (C) \to q,s; (D) \to r$.