Find the distance between the point
(-1, -5, -10) and the point of intersection of the line 
and the plane x – y + z = 5.
Given line,
Let,
⇒ x = 3k+2 , y = 4k-1 ,z = 12k+2
Coordinates of any point on the line are (3k+2 , 4k-1 , 12k+2).
The point of intersection of the line 
and the plane x – y + z = 5 will be in the form (3k+2 , 4k-1 , 12k+2).
we have, plane x – y + z = 5
⇒ 3k+2 – (4k-1)+ 12k+2 =5
⇒ 3k+2 – 4k+1+ 12k+2-5 =0
⇒ 11k = 0
⇒ k = 0
Hence, the point of intersection of the line 
and the plane x – y + z = 5 is (2 , -1 , 2).
The required distance between the (-1, -5, -10) and (2 , -1 , 2)
= 
=
=13 units
Couldn't generate an explanation.
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