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The foot of the perpendicular drawn from the origin to the plane is $(4,-2,-5)$. Hence, the equation of the plane is
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Verified Answer
The correct answer is:
$4 x-2 y-5 z=45$
D.r's of normal to the plane $\langle 4-0,-2-0,-5-0\rangle \equiv\langle 4,-2,-5\rangle$
Hence equation of the plane $4 x-2 y-5 z=\lambda$
But it passes through $(4,-2,-5)$
$\begin{aligned}
& \Rightarrow 4 \times 4-2 \times(-2)-5 \times(-5)=\lambda \\
& \Rightarrow \lambda=45
\end{aligned}$
Hence equation of the plane $4 x-2 y-5 z=\lambda$
But it passes through $(4,-2,-5)$
$\begin{aligned}
& \Rightarrow 4 \times 4-2 \times(-2)-5 \times(-5)=\lambda \\
& \Rightarrow \lambda=45
\end{aligned}$
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