The distance of the point (1, 0, 2) from the point of intersection of the line
x−23=y+14=z−212 and the plane x - y + z = 16 is
13
Given equation of line is
x−23=y+14=z−212=λ [say]...(i)
and equation of plane is
x−y+z=16 .........(ii)
Any point on the line (i) is, (3λ+2,4λ−1,12λ+2)
Let this point of intersection of the line and plane.
∴ (3λ+2)−(4λ−1)+(12λ+2)=16⇒ 11λ+5=16⇒ 11λ=11⇒λ=1
So, the point of intersection is (5, 3, 14).
Now, distance between the points (1, 0, 2) and (5, 3, 14)
=√(5−1)2+(3−0)2+(14−2)2
=√16+9+144=√169=13