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Question
Find the distance of the point P (–1, –5, –10) from the point of intersection of the line joining the points A (2, –1, 2) and B (5, 3, 4) with the plane x – y + z = 5.
Solution
The equation of the line passing through the points A (2,-1,2) and B (5,3,4) is given by
\[\frac{x - 2}{5 - 2} = \frac{y + 1}{3 + 1} = \frac{z - 2}{4 - 2}\]
\[ \Rightarrow \frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{2} = m (\text { say})\]
\[ \Rightarrow x = 3m + 2 , y = 4m - 1 , z = 2m + 2\]
Now, putting the values of x , y and z in the equation of the plane \[x - y + z = 5\],
we get:
\[(3m + 2) - (4m - 1) + (2m + 2) = 5\]
\[ \Rightarrow m = 0\]
So, the point of intersection of the line and the plane is (2,-1,2).
\[\therefore \text { The distance of the point} P ( - 1, - 5, - 10) \text { and the point of intersection} (2, - 1, 2) \text { is }\]
\[ = \sqrt{(3 )^2 + (4 )^2 + (12 )^2} = 13 \text { units }\]
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