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A plane $x$ passes through the point $(1,1,1)$. If $b, c, a$ are the direction ratios of a normal to the plane, where $a, b, c(a < b < c)$ are the factors of 2001 , then the equation of the plane is
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Verified Answer
The correct answer is:
$23 x+29 y+3 z=55$
We have,
$$
\begin{aligned}
& 2001=3.23 .29 \\
& \text { so, } a=3, b=23, c=29
\end{aligned}
$$
Since, direction ratios of a normal are $b, c, a$ then the equation of plane is
$$
\begin{aligned}
b x+c y+a z+d & =0 \\
\Rightarrow \quad 23 x+29 y+3 z+d & =0
\end{aligned}
$$
It passes through $(1,1,1)$
$$
\begin{aligned}
& \therefore \quad 23+29+3+d=0 \\
& \Rightarrow \quad 55+d=0 \\
& \Rightarrow \quad d=-55 \\
&
\end{aligned}
$$
Thus, equation of the plane is
$$
\begin{aligned}
& 23 x+29 y+3 z-55=0 \\
& \Rightarrow \quad 23 x+29 y+3 z=55 \\
&
\end{aligned}
$$
$$
\begin{aligned}
& 2001=3.23 .29 \\
& \text { so, } a=3, b=23, c=29
\end{aligned}
$$
Since, direction ratios of a normal are $b, c, a$ then the equation of plane is
$$
\begin{aligned}
b x+c y+a z+d & =0 \\
\Rightarrow \quad 23 x+29 y+3 z+d & =0
\end{aligned}
$$
It passes through $(1,1,1)$
$$
\begin{aligned}
& \therefore \quad 23+29+3+d=0 \\
& \Rightarrow \quad 55+d=0 \\
& \Rightarrow \quad d=-55 \\
&
\end{aligned}
$$
Thus, equation of the plane is
$$
\begin{aligned}
& 23 x+29 y+3 z-55=0 \\
& \Rightarrow \quad 23 x+29 y+3 z=55 \\
&
\end{aligned}
$$
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