=λ (say) Any point on the line is ‌P(3λ+6,2λ+7,−2λ+7) ‌‌ Let ‌A≡(1,2,3) The d.r.s. of line AP are ‌3λ+6−1,2λ+7−2,−2λ+7−3 ‌‌ i.e. ‌3λ+5,2λ+5,−2λ+4 Since AP is perpendicular to the given line, ‌3(3λ+5)+2(2λ+5)−2(−2λ+4)=0 ‌⇒17λ+17=0 ‌⇒λ=−1 ‌∴‌‌P≡(3,5,9) ‌∴‌‌AP=√(3−1)2+(5−2)2+(9−3)2 ‌=√49 ‌=7‌ units ‌