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A plane $\pi$ passing through the point $3 \bar{i}-4 \bar{j}+5 \bar{k}$ is parallel to the plane which passes through the point $\bar{i}+\bar{j}-\bar{k}$ and perpendicular to the vector $\bar{i}+2 \bar{j}-3 \bar{k}$. Then the cartesian equation of $\pi$ is
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Verified Answer
The correct answer is:
$x+2 y-3 z+20=0$
Equation of plane passing through $\hat{i}+\hat{j}-\hat{k}$ and perpendicular to $\hat{i}+2 \hat{j}-3 \hat{k}$ is
$$
\begin{aligned}
& 1(x-1)+2(y-1)-3(z+1)=0 \\
& x+2 y-3 z-6=0
\end{aligned}
$$
$\because \quad \pi$ plane is parallel to $x+2 y-3 z-6=0$ and passing through $3 \hat{i}-4 \hat{j}+5 \hat{k}$.
$\therefore \quad$ Equation of $\pi$ :
$$
\begin{aligned}
& 1(x-3)+2(y+4)-3(z-5)=0 \\
\Rightarrow \quad & x+2 y-3 z+20=0 .
\end{aligned}
$$
$$
\begin{aligned}
& 1(x-1)+2(y-1)-3(z+1)=0 \\
& x+2 y-3 z-6=0
\end{aligned}
$$
$\because \quad \pi$ plane is parallel to $x+2 y-3 z-6=0$ and passing through $3 \hat{i}-4 \hat{j}+5 \hat{k}$.
$\therefore \quad$ Equation of $\pi$ :
$$
\begin{aligned}
& 1(x-3)+2(y+4)-3(z-5)=0 \\
\Rightarrow \quad & x+2 y-3 z+20=0 .
\end{aligned}
$$
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