Match the statements in column-I with those i | vedclass

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(A-T, B-P, R, C-Q, S, D-R) $(A)-(t)$: Let the line from origin be $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$. It intersects the given lines. Solving the sy

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$A-t, B-p, r, C-q, s, D-r$

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(A-T, B-P, R, C-Q, S, D-R) $(A)-(t)$: Let the line from origin be $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$. It intersects the given lines. Solving the system of equations,we find the intersection points $P(5, -5, 2)$ and $Q(\frac{10}{3}, -\frac{10}{3}, \frac{8}{3})$. The distance $PQ^2 = (5-\frac{10}{3})^2 + (-5+\frac{10}{3})^2 + (2-\frac{8}{3})^2 = (\frac{5}{3})^2 + (-\frac{5}{3})^2 + (-\frac{2}{3})^2 = \frac{25+25+4}{9} = \frac{54}{9} = 6$.$(B)-(p), (r)$: $	an^{-1}(x+3) - 	an^{-1}(x-3) = \sin^{-1}(\frac{3}{5}) = 	an^{-1}(\frac{3}{4})$. Using $	an^{-1}A - 	an^{-1}B = 	an^{-1}(\frac{A-B}{1+AB})$,we get $\frac{(x+3)-(x-3)}{1+(x+3)(x-3)} = \frac{3}{4} \Rightarrow \frac{6}{x^2-8} = \frac{3}{4} \Rightarrow x^2-8 = 8 \Rightarrow x^2 = 16 \Rightarrow x = \pm 4$.$(C)-(q), (s)$: Given $\vec{a} \cdot \vec{b} = 0$,$(\vec{b}-\vec{a}) \cdot (\vec{b}+\vec{c}) = 0$,$2|\vec{b}+\vec{c}| = |\vec{b}-\vec{a}|$,and $\vec{a} = \mu \vec{b} + 4 \vec{c}$. Substituting and simplifying leads to the quadratic equation in $\mu$: $\mu^2 - 5\mu = 0$,giving $\mu = 0, 5$.$(D)-(r)$: $I = \frac{2}{\pi} \int_{-\pi}^{\pi} \frac{\sin(9x/2)}{\sin(x/2)} dx = \frac{4}{\pi} \int_{0}^{\pi} \frac{\sin(9x/2)}{\sin(x/2)} dx$. Using the identity $\frac{\sin(nx)}{\sin x} = 1 + 2\sum_{k=1}^{(n-1)/2} \cos(2kx)$,we get $I = \frac{4}{\pi} \int_{0}^{\pi} (1 + 2\sum_{k=1}^{4} \cos(kx)) dx = \frac{4}{\pi} [x + 2\sum \frac{\sin(kx)}{k}]_0^{\pi} = \frac{4}{\pi} (\pi) = 4$.

column-$I$	column-$II$
$(A)$ $A$ line from the origin meets the lines $\frac{x-2}{1}=\frac{y-1}{-2}=\frac{z+1}{1}$ and $\frac{x-\frac{8}{3}}{2}=\frac{y+3}{-1}=\frac{z-1}{1}$ at $P$ and $Q$ respectively. If length $PQ=d$,then $d^2$ is	$(p)$ $-4$
$(B)$ The values of $x$ satisfying $\tan ^{-1}(x+3)-\tan ^{-1}(x-3)=\sin ^{-1}\left(\frac{3}{5}\right)$ are	$(q)$ $0$
$(C)$ Non-zero vectors $\vec{a}, \vec{b}$ and $\vec{c}$ satisfy $\vec{a} \cdot \vec{b}=0$,$(\vec{b}-\vec{a}) \cdot(\vec{b}+\vec{c})=0$ and $2\|\vec{b}+\vec{c}\|=\|\vec{b}-\vec{a}\|$. If $\vec{a}=\mu \vec{b}+4 \vec{c}$,then the possible values of $\mu$ are	$(r)$ $4$
$(D)$ Let $f$ be the function on $[-\pi, \pi]$ given by $f(0)=9$ and $f(x)=\frac{\sin \left(\frac{9 x}{2}\right)}{\sin \left(\frac{x}{2}\right)}$ for $x \neq 0$. The value of $\frac{2}{\pi} \int_{-\pi}^\pi f(x) dx$ is	$(s)$ $5$
	$(t)$ $6$

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