If the line x/1=y/2=z/3 intersects the t...

If the line `x/1=y/2=z/3` intersects the the line `3beta^(2)+3(1-2alpha)y+z=3-1/2{6alpha^(2)x+3(1-2beta)y+2z}` then point `(alpha,beta,1)` lies on the plane

`2x-y+z=4`

`x+y-z=0`

`x-2y=0`

`2x-y=0`

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The correct Answer is:

A, B, C

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