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If equation of the plane that contains the point (-2,3,5) and is perpendicular to each of the planes(2x+4y+5z)=8 and 3x-2y-3z=5 is \alpha x+\beta y+\gamma z+97=0 then \alpha +\beta +\gamma =

Option: 1

15


Option: 2

18


Option: 3

17


Option: 4

16


Answers (1)

best_answer

The equation of plane through (–2,3,5) is a(x+2)+b(y-3)+c(z-5)=0 it is perpendicular to 2 x+4 y+5 z=8 \& 3 x-2 y+3 z=5 \begin{aligned} & \therefore 2 a+4 b+5 c=0 \\ & 3 a-2 b+3 c=0 \\ & \therefore \frac{a}{\left|\begin{array}{cc} 4 & 5 \\ -2 & 3 \end{array}\right|}=\frac{-b}{\left|\begin{array}{cc} 2 & 5 \\ 3 & 3 \end{array}\right|}=\frac{\mathrm{c}}{\left|\begin{array}{cc} 2 & 4 \\ 3 & -2 \end{array}\right|} \\ & \Rightarrow \frac{a}{22}=\frac{b}{9}=\frac{\mathrm{c}}{-16} \end{aligned} \therefore Equation of plane is

\begin{aligned} & 22(x+2)+9(y-3)-16(z-5)=0 \\ & \Rightarrow 22 x+9 y-16 z+97=0 \\ & \text { Comparing with } \alpha x+\beta y+\gamma x+97=0 \\ & \text { We get } \alpha+\beta+\gamma=22+9-16=15 \end{aligned}

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Suraj Bhandari

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