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For all real values of a and b lines (2a...

For all real values of a and b lines `(2a+b)x + (a+3b)y+(b-3a)=0` and mx+2y+6=0 are concurrent, then m is equal to

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The correct Answer is:
-2
Lines (2a+b)x+(a+3b)y+(b-3a) = 0 or a(2x+y-3) +b(x+3y+1)=0 are concurrent at the point of intersection of lines 2x+y-3 = 0 and x+3y+1 =0, which is (2, -1).
Now, line mx+2y+6= 0 mus pass through this point.
Therefore, 2m-2+6=0 or m=-2.
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