Let L1:x+3∕−3=y−3∕2=z−6∕4=r1 and L2:x−5∕3=y−7∕−1=z+2∕1=r2
Again, let A and B are the points of intersection of the line with L1 and L2. Then, A(−3r1−3,2r1+3,4r1+6) and B(3r2+5,−r2+7,r2−2) Now, DR′s of AB are <3r2+5+3r1+3,−r2+7−2r1−3,r2−2−4r1−6> i.e., <3r2+3r1+8,−r2−2r1+4,r2−4r1−8> But DR′s o f AB are <2,7,−5>. ∴
3r2+3r1+8
2
=
−r2−2r1+4
7
=
r2−4r1−8
−5
⇒7(3r2+3r1+8)=2(−r2−2r1+4) and −5(−r2−2r1+4)=7(r2−4r1−8)
‌⇒23r2+25r1=−48 and −2r2+38r1=−36 ∴r1=−1and r2=−1
So, coordinates of A are (0,1,2) and B are (2,8,−3) ∴ Length of intercept, AB=√(2−0)2+(8−1)2+(−3−2)2 =√4+49+25=√78