MathematicsMedium249×since 2002Q2393The image of the line x−13=y−31=z−4−5 {{x - 1} \over 3} = {{y - 3} \over 1} = {{z - 4} \over { - 5}}\,3x−1=1y−3=−5z−4 in the plane 2x−y+z+3=02x-y+z+3=02x−y+z+3=0 is the line :Ax−33=y+51=z−2−5{{x - 3} \over 3} = {{y + 5} \over 1} = {{z - 2} \over { - 5}}3x−3=1y+5=−5z−2Bx−3−3=y+5−1=z−25 {{x - 3} \over { - 3}} = {{y + 5} \over { - 1}} = {{z - 2} \over 5}\,−3x−3=−1y+5=5z−2Cx+33=y−51=z−2−5 {{x + 3} \over 3} = {{y - 5} \over 1} = {{z - 2} \over { - 5}}\,3x+3=1y−5=−5z−2Dx+3−3=y−5−1=z+25{{x + 3} \over { - 3}} = {{y - 5} \over { - 1}} = {{z + 2} \over 5}−3x+3=−1y−5=5z+2Check answerSkip