The matrix $$ \left|\begin{array}{ccc} a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right| $$ is singular if \(\ldots \ldots\) (a) \(a-b=0\) (b) \(a+b=0\) (c) \(a+b+c=0\) (d) \(a=0\)

We are given the matrix: $$ A = \left|\begin{array}{ccc} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{array}\right| $$

To calculate the determinant, we can use the following formula: $$ \det(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) $$ Plug in the values from the given matrix: $$ \det(A) = (a-b-c)((b-c-a)(c-a-b) - 2b\cdot 2c) - 2a(2b(c-a-b) - 2b\cdot 2c) + 2a(2b\cdot 2c - (b-c-a)2c) $$

To simplify and find when the determinant is zero, expand and factorize the terms: \begin{align*} \det(A) &= (a-b-c)((b^2-c^2-ab-ac+bc) - 4bc) \\ &\quad -2a(-4c+2b+2c-(a-c)(-4c)) \\ &\quad +2a(-4c+2b-(a-b)(-2c)) \\ \end{align*} Now, factorize and combine similar terms: \begin{align*} \det(A) &= (a-b-c)(b^2-c^2-4bc-ab+ac+bc) + 8a^2c-8abc \\ &= (a-b-c)(b^2-5bc+c^2-ab+ac+bc) + 8a^2c-8abc \\ \end{align*} Observe that (b^2-5bc+c^2) can be written as (b-c)^2, and the expression inside the parentheses can be written as a sum of two squares: \begin{align*} \det(A) &= (a-b-c)((b-c)^2 - (a-c)(a-b)) + 8a^2c-8abc \\ \end{align*} The determinant will be zero if the following condition holds: $$ (a-b-c)((b-c)^2 - (a-c)(a-b)) + 8a^2c-8abc = 0 $$

Now, we will compare the condition obtained in Step 3 with the given options: (a) \(a-b=0\): This condition does not match our result. (b) \(a+b=0\): This condition also does not match our result. (c) \(a+b+c=0\): This condition matches the result obtained in Step 3, because if \(a+b+c=0\), then \(a-b-c = 0\) and therefore \(\det(A) = 0\). (d) \(a=0\): This condition does not match our result. Thus, the correct option is (c) \(a+b+c=0\).