The point of intersection of the lines →r=→a+s(→b+→c) , →r=→b+t(→c+→a) is
A
0
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B
a+b
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C
a+b+c
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D
b+c
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Solution
The correct option is Ca+b+c Put s=1 and t=1 and check If s=1, then →r=→a+→b+→c If t=1, then r=→a+→b+→c So the point →a+→b+→c is a point on both the lines.