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The point of intersection of the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z+2}{3}$ and the plane $2 x+3 y+z=0$ is
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Verified Answer
The correct answer is:
$\left(\frac{-1}{11}, \frac{9}{11} \frac{-25}{11}\right)$
$\frac{x}{1}=\frac{y-1}{2}=\frac{z+2}{3}=r \text {, (say) }$
So, $x=r, y=2 r+1, z=3 r-2$
$\begin{aligned}
& \therefore 2 r+3(2 r+1)+(3 r-2)=0 \Rightarrow r=\frac{-1}{11} \\
& x=\frac{-1}{11}, y=\frac{9}{11}, z=-\frac{25}{11} \\
&
\end{aligned}$
So, $x=r, y=2 r+1, z=3 r-2$
$\begin{aligned}
& \therefore 2 r+3(2 r+1)+(3 r-2)=0 \Rightarrow r=\frac{-1}{11} \\
& x=\frac{-1}{11}, y=\frac{9}{11}, z=-\frac{25}{11} \\
&
\end{aligned}$
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