The distance between the line r⃗ .2i^−2j^+3k^+λ(i^−j^+4k^) and the plane r⃗ .(i^−5j^+k^)=5 is

[a] It is obvious that the given line and plane are parallel. Given point on the lie is A(2,−2,3). B(0,0,5) is a point the plane. Therefore, −→−AB=(2−0)i^+(−2−0)j^+(3−5)k^ Then distance of B from the plane = Projection of −→−ABon vector i^+5j^+k^ P=∣∣∣(2i^−2j^−2k^).(i^+5j^+k^)1+25+1−−−−−−−−√∣∣∣=∣∣∣2−10−227−−√∣∣∣=1033√