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The distance of the point (1, 3, −7) from the plane passing through the point (1, −1, −1), having normal perpendicular to both the lines (x−1)/1 = (y+2)/(−2) = (z−4)/3 and (x−2)/2 = (y+1)/(−1) = (z+7)/(−1), is:

1. 10/√74
2. 20/√74
3. 10/√83
4. 5/√83

Correct answer: 10/√83

Solution

The normal perpendicular to both lines is (1,-2,3)x(2,-1,-1) = (5,7,3). Plane through (1,-1,-1): 5x+7y+3z+5=0. Distance of (1,3,-7) = |5+21-21+5|/sqrt(25+49+9) = 10/sqrt(83).

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