if m =(2^x)*(5^y)*(7^z)

If both 350 and 280 are factors of m, m should be able to accomodate all the powers of 2,5 and 7.
Since we have to find the minimum value of x,y and z we will pick the highest powers of 2,5 and 7 from 350 and 280.

350 = (2^1)*(5^2)*(7^1)
280 = (2^3)*(5^1)*(7^1)

max x = 3
max y = 2
max z = 1

so minimum value xyz = 6

[spoiler](D)[/spoiler], IMO.

Let me try to validate this answer.
I am claiming that minimum m = (2^3)*(5^2)*(7^1) = 1400
Are both 350 and 280 factors of m?
YES!

Check OA.

Me too got 6 .... but the OA given in B ... no explanation given ... guess 6 is correct then