If the line, \(\dfrac{x-3}{1}=\dfrac{y+2}{-1}=\dfrac{z+\lambda}{-2}\) lies on the plane \(2x-4y+3z=2\), then the shortest distance between this line and the line \(\dfrac{x-1}{12}=\dfrac{y}{9}=\dfrac{z}{4}\) is

1
\(0\)
2
\(2\)
3
\(1\)
4
\(3\)

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