If the shortest distance between the lines x-12= y-23= z-3 and x-21= y-44= z-55 is 1 3, the sum of possible value of is

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Correct answer: A. (a) : The shortest distance between two lines l_1 : x-x_1a_1= y-y_1b_1= z-z_1c_1 l_2 : x-x_2a_2= y-y_2b_2= z-z_2c_2 is given by ( (x_2-x_1y_2-y_1z_2-z_1a_1b_1c_1a_2b_2c_2) (a_1b_2-a_2b_1)^2+(b_1c_2-b_2c_1)^2+(c_1a_2-c_2a_1)^2 ) So, shortest distance between the lines x-12= y-23= z-3 and x-21= y-44= z-55 = ( (12223 145 ) (5 )^2+ (15-4 )^2+ ( -10 )^2 )= 1 3 ⇒ ( (15-4 )-2 (10- )+2 5 25+ (15-4 )^2+ ( -10 )^2 )= 1 3 ⇒ ( 15-4 -20+2 +10 25+225+16 ^2-120 + ^2+100-20 )= 1 3 ⇒ ( 5-2 350+17 ^2-140 )= 1 3 ⇒ (5-2 )^2350+17 ^2-140 = 13 ⇒3(25+4 ^2-20 )=350+17 ^2-140 ⇒75+12 ^2-60 =350+17 ^2-140 ⇒5 ^2-80 +275=0 ⇒ ^2-16 +55=0 ⇒ ^2-11 -5 +55=0 ⇒ ( -11 )-5 ( -11 )=0 ⇒ ( -5 ) ( -11 )=0 ⇒ =5 ,11 Sum of all values of =5+11=16

$If the shortest distance between the lines \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{λ} and \frac{x - 2}{1} = \frac{y - 4}{4} = \frac{z - 5}{5} is \frac{1}{3}, the sum of possible value of λ is$

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