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Let the mirror image of the point $(a, b, c)$ with respect to the plane $3 x-4 y+12 z+19=0$ be $(a-6, \beta, \gamma)$ . If $a+b+c=5$, then $7 \beta-9 \gamma$ is equal to______________.
Option: 1
137

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

$\mathrm{\frac{x-a}{3}=\frac{y-b}{-4}=\frac{3-c}{12}=-\frac{2(3 a-4 b+12 c+19)}{3^{2}+4^{2}+12^{2}}}$

$\mathrm{\text { Here } x=a-6, y=\beta, z=\gamma} . \\$

$\mathrm{\Rightarrow \frac{a-6-a}{3}=\frac{\beta-b}{-4}=\frac{\gamma-c}{12}=\frac{-2(3 a-4 b+12 c+19)}{169}}$

$\mathrm{\Rightarrow 3 a-4 b+12 c =150 }\\$ .............(1)

$\mathrm{a+b+c =5} \\$

$\mathrm{\Rightarrow 3 a+3 b+3 c =15}$ .................(2)

Subtracting Eqn (1)-(2)

$\mathrm{ \Rightarrow 7 b-9 c=-135}$ .................(3)

$\mathrm{\frac{\beta-b}{-4}=-2 \Rightarrow b=\beta-8,} \\$

$\mathrm{\frac{\gamma-c}{12}=-2 \Rightarrow c=\gamma+24, }$

Substituting in (3)

$\mathrm{7(\beta-8) -9(\gamma+24)=-135} \\$

$\mathrm{\Rightarrow 7 \beta-9 \gamma =56+216-135 }\\$

$\mathrm{=272-135=137}$

So answer is $\mathrm{137}$

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Kshitij

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