The sum of the coordinates of one of the points on the line x - 2 1 = y + 3 -2 = z + 5 2 which is at a distance of 3 units from the point (2, -3, -5) is ...

Step-by-step solution

Let any point on the given line x - 2 1 = y + 3 -2 = z + 5 2 = be (2 + , -3 - 2 , -5 + 2 ) . The distance of this point from (2, -3, -5) is given as 3 units. Using the distance formula: (2 + - 2)^2 + (-3 - 2 + 3)^2 + (-5 + 2 + 5)^2 = 3 ^2 + (-2 )^2 + (2 )^2 = 3 ^2 + 4 ^2 + 4 ^2 = 3 9 ^2 = 3 3| | = 3 = 1 For = 1 , the coordinates of the point are (2 + 1, -3 - 2(1), -5 + 2(1)) = (3, -5, -3) . The sum of the coordinates is 3 - 5 - 3 = -5 . For = -1 , the coordinates of the point are (2 - 1, -3 - 2(-1), -5 + 2(-1)) = (