Let Q(3λ+6,2λ+7,−2λ+7) be any point lie on line
x−6
3
=
y−7
2
=
z−7
−2
=λ
x−6
3
=
y−7
2
=
z−7
−2
∴ Direction ratio of line PQ(3λ+5,2λ+5,−2λ+4) Since, line PQ is perpendicular to given line. ∴3(3λ+5)+2(2λ+5)−2(−2λ+4)=0 ⇒17λ+17=0⇒λ=−1 ∴Q(3,5,9) Distance (PQ)=√(3−1)2+(5−2)2+(9−3)2 =√4+9+36 =√49=7