 Good morning friends. I am Purva and today we will work out the following question. Find the shortest distance between the lines. x plus 1 upon 7 is equal to y plus 1 upon minus 6 is equal to z plus 1 upon 1 and x minus 3 upon 1 is equal to y minus 5 upon minus 2 is equal to z minus 7 upon 1. Let the equation of the line L1B x minus x1 upon a1 is equal to y minus y1 upon b1 is equal to z minus z1 upon c1 and equation of the line L2B x minus x2 upon a2 is equal to y minus y2 upon b2 is equal to z minus z2 upon c2. Then the shortest distance between the lines is given by D is equal to determinant of x2 minus x1 y2 minus y1 z2 minus z1 and below x2 minus x1 we write a1 below y2 minus y1 we write b1 and below z2 minus z1 we write c1 below a1 we write a2 below b1 we write b2 below c1 we write c2 upon under root of b1 c2 minus b2 c1 whole square plus c1 a2 minus c2 a1 whole square plus a1 b2 minus a2 b1 whole square. So this is the key idea behind our question. Let us now begin with the solution. Now we are given equation of the line L1 as x plus 1 upon 7 is equal to y plus 1 upon minus 6 is equal to z plus 1 upon 1 and equation of line L2 is x minus 3 upon 1 is equal to y minus 5 upon minus 2 is equal to z minus 7 upon 1. Now determinant of x2 minus x1 y2 minus y1 z2 minus z1 a1 b1 c1 a2 b2 c2 is equal to determinant of Now by comparing the equation of the lines L1 and L2 with the equation of the lines L1 and L2 given in key idea we can clearly see that here x2 is equal to 3 and x1 is equal to minus 1 so we get x2 minus x1 is equal to 4 y2 is equal to 5 and y1 is equal to minus 1 so we get y2 minus y1 is equal to 6 and z2 is equal to 7 and z1 is equal to minus 1 so we get z2 minus z1 is equal to 8 here a1 b1 and c1 are 7 minus 6 and 1 respectively so we get 7 minus 6 1 and a2 b2 and c2 are 1 minus 2 and 1 now this is equal to 4 into minus 6 into 1 is minus 6 minus minus 2 into 1 is minus 2 minus into minus becomes plus so we get plus 2 minus 6 into 7 into 1 is 7 minus 1 into 1 is 1 plus 8 into 7 into minus 2 is minus 14 minus 1 into minus 6 is minus 6 minus into minus becomes plus so we get plus 6 this is equal to 4 into minus 6 plus 2 is minus 4 minus 6 into 7 minus 1 is 6 plus 8 into minus 14 plus 6 is minus 8 this is equal to minus 16 minus 36 minus 64 and this is equal to minus 116 we mark this as 1 also under root of b1 c2 minus b2 c1 whole square plus c1 a2 minus c2 a1 whole square plus a1 b2 minus a2 b1 whole square is equal to under root of b1 c2 now b1 is equal to minus 6 and c2 is equal to minus 1 so we get b1 c2 is equal to minus 6 minus b2 c1 is equal to minus 2 into 1 that is minus 2 minus into minus becomes plus so we get plus 2 whole square plus c1 a2 that is 1 into 1 which is equal to 1 minus c2 a1 that is 1 into 7 which is equal to 7 whole square plus a1 b2 that is 7 into minus 2 which is equal to minus 14 minus a2 b1 that is 1 into minus 6 which is equal to minus 6 minus into minus becomes plus so we get plus 6 whole square this is equal to under root of minus 6 plus 2 is equal to minus 4 and minus 4 whole square is equal to 16 plus 1 minus 7 is equal to minus 6 and minus 6 whole square is equal to 36 plus minus 14 plus 6 is equal to minus 8 and minus 8 whole square is equal to 64 this is equal to under root of 116 which is further equal to 2 under root 29 we mark this as 2 now by key idea we know that the shortest distance is given by d is equal to determinant of x2 minus x1 y2 minus y1 z2 minus z1 a1 b1 c1 a2 b2 c2 upon under root of b1 c2 minus b2 c1 whole square plus c1 a2 minus c2 a1 whole square plus a1 b2 minus a2 b1 whole square now putting the value of 1 and 2 in this equation we get d is equal to minus 116 upon 2 under root 29 now since distance can't be negative therefore we get d is equal to 116 upon 2 under root 29 and we get this is equal to 2 under root 29 thus we have got our answer as 2 under root 29 hope you have understood the solution bye and take care