find the distance of the point (1,-2,3) from the plane x-y+z=5 measured along a line parellel to x/2=y/3=z/-6

The given plane is x – y + z = 5 ... (1)

We have to find the distance of point (1, –2, 3) from this plane measured along a line 11 to

So, the direction ratio of the line from the point (1, –2, 3) to the given plane will be same as that of given line.

Let this line from point (1, –2, 3) meets at Q on the plane.

Equation of line passing through (1, –2, 3) and having D.R's (2, 3, –6) is

∴ Co-ordinates of any point on the line are 2λ + 1,  3λ – 2, –6λ + 3.Q lies on the line

Q lies on the line, therefore, co-ordinate of Q are (2λ + 1, 3λ – 2, –6λ + 3) for some λ.

But, Q also lies in the plane, as it is point of intersection of line and plane. So it must satisfy (1).

∴ Co-ordinates of Q are i.e.,

Using distance formula, we have