Find the coordinates of the foot of the perpendicular drawn from the origin to the plane `2x - 3y + 4z - 6 = 0`.

Explanation

The given plane is,

`2x-3y+4z-6=0`

or, `2x-3y+4z=6`

From the above, it can be deduced that its normal vector from the origin is,

`vecN=2hati-3hatj+4hatk`

So, `abs(vecN)=sqrt(2^2+(-3)^2+4^2)`

`=>abs(vecN)=sqrt(4+9+16)`

`=>abs(vecN)=sqrt(29)`

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