Image of the line x−21=y−12=z3 in the plane 2x+y+z=5 is the line-
A
x−2−1=y−1−2=z15
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B
x−211=1−y1=z−2
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C
x−211=y−11=z2
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D
x−211=1−y1=z2
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Solution
The correct option is Ax−2−1=y−1−2=z15 Bcoordinates(2,1,0)Acoordinates(3,3,3)Awillbecalculatedofx−32=y−31=2−11=t[2t+3,t+3,t+1]⋅[1,2,3]=02t+32t+6+3t+3=07t+12=0t=−127A′[−37,97,−57]Willbex−2−1=y−1−2=z15