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If $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are the intercepts made by the plane passing through the point $(1,2,3)$ parallel to the plane $3 \mathrm{x}+4 \mathrm{y}-5 \mathrm{z}=0$ on $\mathrm{X}, \mathrm{Y}, \mathrm{Z}$-axes respectively then $3 \mathrm{a}+\mathrm{b}+5 \mathrm{c}=$
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The correct answer is:
-1
Since a,b,c are intercepts made by the plane passin through the point $(1,2,3)$
$\Rightarrow$ Let $P_1=\frac{x}{a}+\frac{4}{b}+\frac{z}{c}=1$
Let $P_2=3 x+4 y-5 z=0$
$\because \mathrm{P}_2$ II $\mathrm{P}_1$ and $\mathrm{P}_1$ passing the $(1,2,3)$ now $\mathrm{P}_1$ can be written as
$$
\begin{aligned}
& P_1=3(x-1)+4(y-2)-5(z-3)=0 \\
& P_1=3 x+4 y-5 z+4=0 \\
& P_1=\frac{x}{\left(\frac{-4}{3}\right)}+\frac{y}{(-1)}+\frac{z}{\left(\frac{4}{5}\right)}=1
\end{aligned}
$$
on comparing (i) and (iii)
$$
\begin{aligned}
& \Rightarrow \mathrm{a}=\frac{-4}{3}, \mathrm{~b}=-1, \mathrm{c}=\frac{4}{5} \\
& \Rightarrow 3 \mathrm{a}+\mathrm{b}+5 \mathrm{c}=-1
\end{aligned}
$$
$\Rightarrow$ Let $P_1=\frac{x}{a}+\frac{4}{b}+\frac{z}{c}=1$
Let $P_2=3 x+4 y-5 z=0$
$\because \mathrm{P}_2$ II $\mathrm{P}_1$ and $\mathrm{P}_1$ passing the $(1,2,3)$ now $\mathrm{P}_1$ can be written as
$$
\begin{aligned}
& P_1=3(x-1)+4(y-2)-5(z-3)=0 \\
& P_1=3 x+4 y-5 z+4=0 \\
& P_1=\frac{x}{\left(\frac{-4}{3}\right)}+\frac{y}{(-1)}+\frac{z}{\left(\frac{4}{5}\right)}=1
\end{aligned}
$$
on comparing (i) and (iii)
$$
\begin{aligned}
& \Rightarrow \mathrm{a}=\frac{-4}{3}, \mathrm{~b}=-1, \mathrm{c}=\frac{4}{5} \\
& \Rightarrow 3 \mathrm{a}+\mathrm{b}+5 \mathrm{c}=-1
\end{aligned}
$$
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