Point (3,−2,−λ) on p line 2x−4y+3z−2=0=6+8−3λ−2=0=3λ=12 λ=4 Now,
x−3
1
=
y+2
−1
=
z+4
−2
=k1 ......(i)
x−1
12
=
y
9
=
z
4
=k2 .......(ii) Point on equation (i) P(k1+3,−k1−2,−2k1−4) Point on equation (ii) Q(12k2+1,9k2,4k2) k1+3=12k2+1|−k1−2=9k2|−2k1−4=4k2 k2=0 k1=−2 p(1,0,0) lie on equation of a line 1 gives shortest distance =0