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Reduce the equation of the plane 2x+3y-z...

Reduce the equation of the plane `2x+3y-z=6` to intercept form and find its intercepts on the coordinate axes.

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The given plane is `2x+3y-z=6`
We can write this equation as
`x/3+y/2+z/(-6)=1 ldots`(i)
Then the equation of plane whose intercepts on the coordinate axes are a,b,c is
`x/a+y/b+z/c=1 ldots`(ii)
On comparing (i) and (ii) we get
`a=3,b=2,c=-6`
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