The distance of the point (1, 2, 3) from the line given by the equations x + y = 13 and x - z = -10 , measured parallel to the line x 2 = y 3 = z 6 , is :

Step-by-step solution

Let the line passing through the point P(1, 2, 3) and parallel to x 2 = y 3 = z 6 be L . The parametric equations of line L are: x = 2t + 1 y = 3t + 2 z = 6t + 3 Let this line intersect the given line at point Q . Since Q lies on the line given by x + y = 13 and x - z = -10 , its coordinates must satisfy both plane equations. Substituting the parametric coordinates into the first plane equation x + y = 13 : (2t + 1) + (3t + 2) = 13 5t + 3 = 13 5t = 10 t = 2 We can verify this with the second plane equation x - z =