# Find the distance between the point (−1, −5, −10) and the point of intersection of the line (x-2)/3=(y+1)/4=(z-2)/12 and the plane x-y+z=5 - Mathematics

Find the distance between the point (−1, −5, −10) and the point of intersection of the line (x-2)/3=(y+1)/4=(z-2)/12 and the plane x-y+z=5

#### Solution

Let (x-2)/3=(y+1)/4=(z-2)/12 =lambda

=>x=3lambda+2, y=4lambda-1, z=12lambda+2

The coordinates of any point on the line are given by (3λ+2, 4λ1, 12λ+2).

The point of intersection of the line (x-2)/3=(y+1)/4=(z-2)/12 and the plane xy + z = 5 will also be in the form (3λ+2, 4λ1, 12λ+2) and it will satisfy the equation of plane.

Putting x=3λ+2, y=4λ1 and z= 12λ+2 in xy + z = 5, we have

3λ+2(4λ1)+12λ+2=5

11λ+5=5

11λ=0

λ=0

∴ x = 2, y = −1, z = 2

Hence, the point of intersection of
(x-2)/3=(y+1)/4=(z-2)/12 and the plane xy + z = 5 is (2, −1, 2).

∴ Distance between the point (−1, −5, −10) and (2, −1, 2)

=sqrt((2+1)^2+(−1+5)^2+(2+10)^2)        [Using distance formula]

=sqrt(3^2+4^2+12^2)

= sqrt169

=13

Thus, the distance between the point (−1, −5, −10) and the point of intersection of the line (x-2)/3=(y+1)/4=(z-2)/12

and the plane xy + z = 5 is 13 units.

Concept: Three - Dimensional Geometry Examples and Solutions
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