The distance of the point (1, 0, –3) from the plane x – y – z = 9 measured parallel to the line 
= 
= 
is -
Text Solution
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Given plane is x – y – z = 9 … (1)
Given line AB is 
= 
= 
… (2)
Equation of line passing through (1, 0, –3) and parallel to
= 
= 
is:
= 
= 
= r … (3)
Co-ordinates of any point on (3) may be given as
p (2r + 1, 3r, –6r – 3)
If P is intersection of (1) and (3) then it must lie on (1) ;
(2r + 1) – (3r) – (–6r – 3) = 9
2r + 1 – 3r + 6r + 3 = 9 ⇒ r = 1
∴ co-ordinate of P are (3, 3, –9)
∴ Required distance = distance between (1, 0, –3) and
(3, 3, –9) = 7.
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