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If the points $(1,1, \lambda)$ and $(-3,0,1)$ are equidistant from the plane
$3 x+4 y-12 z+13=0$, then integer value of $\lambda$ is
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$3 x+4 y-12 z+13=0$, then integer value of $\lambda$ is
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Verified Answer
The correct answer is:
$1$
Given $\mathrm{A}(1,1, \lambda)$ and $\mathrm{B}(-3,0,1)$ are equidistant from $3 \mathrm{x}+4 \mathrm{y}-12 \mathrm{z}+13=0$
$$
\begin{aligned}
& \therefore\left|\frac{3(1)+4(1)-12 \lambda+13}{\sqrt{9+16+144}}\right| \\
& \therefore\left|\frac{20-12 \lambda}{13}\right|=\left|\frac{-8}{13}\right| \\
\therefore & 20-12 \lambda=\pm 8 \\
\therefore & 20-12 \lambda=8 \text { or } 20-12 \lambda=-8 \Rightarrow \lambda=1 \text { or } \lambda=\frac{7}{3}
\end{aligned}
$$
$$
\begin{aligned}
& \therefore\left|\frac{3(1)+4(1)-12 \lambda+13}{\sqrt{9+16+144}}\right| \\
& \therefore\left|\frac{20-12 \lambda}{13}\right|=\left|\frac{-8}{13}\right| \\
\therefore & 20-12 \lambda=\pm 8 \\
\therefore & 20-12 \lambda=8 \text { or } 20-12 \lambda=-8 \Rightarrow \lambda=1 \text { or } \lambda=\frac{7}{3}
\end{aligned}
$$
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