Find the equations of the planes parallel to the plane `x - 2y + 2z -4 = 0`, which are at a unit distance from the point (1, 2, 3).

Explanation

Any plane || to the palne ` x - 2y + 2z - 4 = 0` can be written as :
` x - 2y + 2z - lambda = 0` ….(i)
where ` lambda` is a scalar quantity
Now , distance of the from the point (1,2,3)
` = pm (1 (1) - 2 (2) + 2 (3) + lambda)/(sqrt(1^(2)+(-2)^(2)+2^(2)))`
` = pm (1 - 4 + 6 + lambda)/(sqrt(1 + 4 + 4))`
` = pm (3 +lambda)/sqrt9 = pm (3 + lambda)/3`
But `pm (3 + lambda)/3 = 1`
or ` 3 + lambda = pm 3 `
or ` lambda = 0, -6`
Substituting the values of `lambda` in (i) , we have
` x - 2y + 2z - (-6) = 0`
or ` x - 2y + 2z + 6 = 0`