Answer :

**Given,** A line at a distance of 5 units from the point P(1, 3, 3)

**To find :** Find the points on the given line.

**Explanation:** We have line given

Then, x = 3μ - 2 , y = 2μ - 1 and z = 2μ + 3

So, The points on the given line are

(x, y, z) = ( 3μ - 2 , 2μ - 1 , 2μ + 3)

Now, By using distance formula between two points

Therefore, The points are (3(0) - 2 , 2(0) - 1 , 2(0) + 3) and (3(2) - 2 , 2(2) - 1 , 2(2) + 3)

**Hence, Points are ( - 2, - 1, 3) and (4, 3, 7).**

**OR**

Let the equation of the plane passing through A(3, - 1, 2) is A(x - 3) + B(y + 1) + C(z - 2)

And, It also passes through the points B(5, 2, 4) and C( - 1, - 1, 6)

Then, Take point B and equation will be

A(5 - 3) + B(2 - ( - 1)) + C(4 - 2) = 0

So, 2A + 3B + 2C = 0 - - - - - - (i)

And, From point C , the equation will be

A( - 1 - 3) + B( - 1 - ( - 1)) + C(6 - 2) = 0

So, - 4A + 0B + 4C = 0 - - - - - (ii)

On Solving equation (i) and (ii), we get

Now, The Equation of plane is given by

3(x - 3) - 4(y + 1) + 3(z - 2) = 0

3x - 4y + 3z = 19

And, Distance of point P(6, 5, 9) from the given plane

**Hence, The distance from the given plane is**

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