The symmetric form of the line of intersection of two planes x−y+2z=5,3x+y+z=7 is

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4x−11−3=4y+93=z1

x−11−3=y+93=z1

4x−13=4y−93=z1

x−33=y+2−5=z−4

answer is D.

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Detailed Solution

The given planes are x−y+2z=5,3x+y+z=7Consider the normal vectors and then find the cross product of those two vectors to get the vector along the line of intersection of the given two planesn1×n2=ijk1−12311=i(−3)−j(−5)+k(4)Hence the direction ratios of the required line are proportional to ⟨3,−5,−4⟩To get the point of intersection, plug in z=0 in the two plane equations and then solve for x,y x−y=5,3x+y=7Adding these two equations and then simplify x=3,y=-2Therefore, the equation of the line is x−33=y+2−5=z−4

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