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The line of shortest distance between the lines $\mathrm{\frac{x-2}{0}=\frac{y-1}{1}=\frac{z}{1}}$ and $\mathrm{\frac{x-3}{2}=\frac{y-5}{2}=\frac{z-1}{1}}$ makes an angle of $\mathrm{\cos ^{-1}\left(\sqrt{\frac{2}{27}}\right)}$ with the plane $\mathrm{\mathrm{P}: \mathrm{a} x-y-z=0}$ , $\mathrm{(a>0)}$ . If the image of the point $\mathrm{(1,1,-5)}$ in the plane $\mathrm{P}$ is $\mathrm{(\alpha, \beta, \gamma)}$ , then $\mathrm{\alpha+\beta-\gamma}$ is equal to ________.
Option: 1
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Option: 2
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Option: 3
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Option: 4
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Answers (1)

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DR's of line of shortest distance

$\mathrm{\left|\begin{array}{ccc} \hat{i} \cdot \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ 2 & 2 & 1 \end{array}\right|=-\hat{i}+2 \hat{J}-2 k^{r}}$

angle between line and plane is $\mathrm{\cos ^{-1} \sqrt{\frac{2}{27}}=\alpha}$

$\mathrm{\cos \alpha=\sqrt{\frac{2}{27}}, \quad \sin \alpha=\frac{5}{3 \sqrt{3}}}$

DR's normal to plane $(1,-1,-1)$

$\sin \alpha=\left|\frac{-a-2+2}{\sqrt{4+4+1} \sqrt{a^{2}+1+1}}\right|=\frac{5}{3 \sqrt{3}} \\$

$\mathrm{\sqrt{3}|a|=5 \sqrt{a^{2}+2}} \\$

$\mathrm{3 a^{2}=25 a^{2}+50} \\$

$\text { No value of } a$

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