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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

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#### 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, 4λ−1, 12λ+2).

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

Putting x=3λ+2, y=4λ−1 and z= 12λ+2 in *x* − *y* + *z* = 5, we have

3λ+2−(4λ−1)+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 *x* − *y* + *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 *x* − *y* + *z* = 5 is 13 units.

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