A line ‘l’ passing through origin is perpendicular to the lines             l1:​r→ =(3+t)i^+(−1+2t)j^+(4+2t)k^l2:​ r→ =(3+2s)i^+(3+2s)j^+(2+s)k^                        If the coordinates of the point in the first octant on ‘’l2 at a distance of17 from the point of intersection of ‘l’ and l1‘’ are (a, b, c) then18(a+b+c) is equal to…..

The given lines are L1≅x−31=y+12=z−42  and L2≅x−32=y−32=z−21The direction ratios of line which is perpendicular to the above tow lines isijk122221=−2i+3j−2k            Hence, the equation of the line perpendicular to the above lines and passing through the origin is L≃x−2=y3=z−2            General point on the above line is−2k,3k,−2kIf it lies on the line 1, then −2k−31=3k+12=−2k−42            Hence,                            3k+1=−2k−45k=−5k=−1                                 The point of intersection of L and first line is P2,−3,2            The point Q on the second line is 2m+3,2m+3,m+2            The distance from P and Q isPQ=17            It gives                           2m+12+2m+62+m2=174m2+4m+1+4m2+24m+36+m2=179m2+28m+20=09m2+10m+18m+20=0m9m+10+29m+10=0m+29m+10=0                                  It givesm=−2,m=−109            As Q lies in the first octant then Q79,79,89            Hence,18a+b+c=44