If the distance between the planes Ax-2y+z=d and the plane containing the lines (x-1)/(2)=(y-2)/(3)=(z-3)/(4) and (x-2)/(3)=(y-3)/(4)=(z-4)/(5)is sqrt(6), then |d| is :

If the distance between the planes Ax-2y+z=d and the plane containing

1.	If the distance between the planes Ax-2y+z=d and the plane containing the lines (x-1)/(2)=(y-2)/(3)=(z-3)/(4) and (x-2)/(3)=(y-3)/(4)=(z-4)/(5)is sqrt(6), then \|d\| is :
Answer» 3 4 5 6 Solution :Equation of the PLANE containingthe GIVEN lines is. `\|{:(x-1,y-2,z-3,),(2,3,4,),(3,4,5,):}\|=0impliesx-2y+z=0` since this plane MUST be parallel to the plane `Ax-2y+z=d`, we GET `A=1`As distance between the PLANES is `sqrt(6)`. we get `(\|d\|)/(sqrt(1+4+1))sqrt(6)implies\|d\|=6`

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If the distance between the planes Ax-2y+z=d and the plane containing the lines (x-1)/(2)=(y-2)/(3)=(z-3)/(4) and (x-2)/(3)=(y-3)/(4)=(z-4)/(5)is sqrt(6), then |d| is :

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