The line (x-2)/(3)=(y+1)/(2)=(z-1)/(-1) ...

The line `(x-2)/(3)=(y+1)/(2)=(z-1)/(-1)` intersects the curve `x^2+y^2=r^2, z=0`, then

Equation of the following through (0, 0, 0) perpendicular to the given line is `3x+2y-z=0`

`r=sqrt(26)`

`r=6`

`r=7`

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(a, b)

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