Difference of normals exceeding a third normal

Let $X$, $Y$, $Z$ be independent normal random variables, each with mean $12$ and variance $12$. The probability that $X-Y$ is larger than $Z$ can be expressed as $\Phi(a)$, where $\Phi$ is the standard normal CDF. Find $a$.

$a=-2$. $W=X-Y-Z$ is normal with mean $12-12-12=-12$ and variance $3\cdot12=36$. Then $P(W>0)=P\!\left(\tfrac{W+12}{6}>2\right)=1-\Phi(2)=\Phi(-2)$.