Without directly evaluting, show that the determinant of $A = 0$ 
Without directly evaluting, show that 
  $det \left[
       \begin{array}{ccc}
         b + c & c + a & b + a \\
           a & b & c \\
           1 & 1 & 1 
       \end{array}
     \right]
=0$  

I am stuck on this one. I can only do this by evuating.
Things that I know:
1) Square matrix $A$ with two proportional rows or columns as a $det(A) = 0$
2) Square matrix $A$ has $det(A)=0$ if it has a row or column of zeros.
Can anyone help?
Thanks.

The row vectors are linearly dependent.  
Specifically, denoting the row vectors as $$\vec r_1=(b+c,a+c,a+b)\quad \vec r_2=(a,b,c)\quad \vec r_3=(1,1,1)$$  then we have $$\vec r_1+\vec r_2=(a+b+c)\,\vec r_3$$
