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A plane passes through the point $(3,5,7)$. If the direction ratios of its normal are equal to the intercepts made by the plane $x+3 y+2 z=9$ with the coordinate axes, then the equation of that plane is
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Verified Answer
The correct answer is:
6x + 2 y + 3z = 49
Given equation of plane.
$$
\begin{aligned}
x+3 y+2 z & =9 \\
\Rightarrow \quad \frac{x}{9}+\frac{y}{3}+\frac{z}{\left(\frac{9}{2}\right)} & =1
\end{aligned}
$$
$\Rightarrow$ So, intercepts are $9,3, \frac{9}{2}$.
According to the questions,
direction ratio of normal is $9,3, \frac{9}{2}$ and plane passes through $(3,5,7)$
$\therefore$ Equation of plane is
$$
\begin{aligned}
& & (x-3) 9+(y-5) 3+(z-7) \frac{9}{2} & =0 \\
& \Rightarrow & 9 x-27+3 y-15+\frac{9}{2} z-\frac{63}{2} & =0 \\
\Rightarrow & & 18 x+6 y+9 z & =54+30+63 \\
\Rightarrow & & 18 x+6 y+9 z & =147 \\
& & 6 x+2 y+3 z & =49
\end{aligned}
$$
$$
\begin{aligned}
x+3 y+2 z & =9 \\
\Rightarrow \quad \frac{x}{9}+\frac{y}{3}+\frac{z}{\left(\frac{9}{2}\right)} & =1
\end{aligned}
$$
$\Rightarrow$ So, intercepts are $9,3, \frac{9}{2}$.
According to the questions,
direction ratio of normal is $9,3, \frac{9}{2}$ and plane passes through $(3,5,7)$
$\therefore$ Equation of plane is
$$
\begin{aligned}
& & (x-3) 9+(y-5) 3+(z-7) \frac{9}{2} & =0 \\
& \Rightarrow & 9 x-27+3 y-15+\frac{9}{2} z-\frac{63}{2} & =0 \\
\Rightarrow & & 18 x+6 y+9 z & =54+30+63 \\
\Rightarrow & & 18 x+6 y+9 z & =147 \\
& & 6 x+2 y+3 z & =49
\end{aligned}
$$
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