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The shortest distance between the two lines L1: 3(x-1) = 6(y-2) = 2(z-1) and L2: 4(x-2) = 2(y-lambda) = (z-3), where lambda is a real number, equals 1/sqrt(38). Find the integral (integer) value of lambda.

1. 3
2. 2
3. 5
4. -1

Correct answer: 3

Solution

The cross product d1 x d2 = (-2, -5, 3) with magnitude sqrt(38). The vector connecting reference points P1=(1,2,1) and P2=(2,lambda,3) is (1, lambda-2, 2). Setting |(1,lambda-2,2).(-2,-5,3)| = 1 gives |14-5*lambda| = 1, yielding lambda = 3 or lambda = 13/5. The integer value is 3.

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