If $X$ and $Y$ are independent random variable following the normal distribution $N(\mu, \sigma^2)$ with mean $\mu$ and variance $\sigma^2$ such that $X \sim N(\mu_{X}, \sigma_{X}^2)$ and $Y \sim N(\mu_{Y}, \sigma_{Y}^2)$, and if $T$ is some constant, then the probability $P(X-Y<T)$
$$P(Z<T)=\Phi\left(\frac{T-\mu_Z}{\sigma_Z}\right),$$
where $Z =X-Y$, $ \mu_{Z}=\mu_{X}-\mu_{Y}=$, and $ \sigma_{Z}^2=\sigma_{X}^2 + \sigma_{Y}^2$. (The detailed explanations can be found here.)
I want to extend this to the case of random vectors $\mathbf{X}$ and $\mathbf{Y}$. Let us consider the random vectors of length $L=3$.
Then, for $X_1$ conditioned on the $\mathbf{Y}$
$$P(X_1-Y_1<T)=\Phi\left(\frac{T-\mu_{X_1}-\mu_{Y_1}}{\sigma^2_{X_1}+\sigma^2_{Y_1}}\right),$$ $$P(X_1-Y_2<T)=\Phi\left(\frac{T-\mu_{X_1}-\mu_{Y_2}}{\sigma^2_{X_1}+\sigma^2_{Y_2}}\right),$$ $$P(X_1-Y_3<T)=\Phi\left(\frac{T-\mu_{X_1}-\mu_{Y_3}}{\sigma^2_{X_1}+\sigma^2_{Y_3}}\right),$$
and for $X_2$ conditioned on the $\mathbf{Y}$ $$P(X_2-Y_1<T)=\Phi\left(\frac{T-\mu_{X_1}-\mu_{Y_1}}{\sigma^2_{X_2}+\sigma^2_{Y_1}}\right),$$ $$P(X_2-Y_2<T)=\Phi\left(\frac{T-\mu_{X_1}-\mu_{Y_2}}{\sigma^2_{X_2}+\sigma^2_{Y_2}}\right),$$ $$P(X_2-Y_3<T)=\Phi\left(\frac{T-\mu_{X_1}-\mu_{Y_3}}{\sigma^2_{X_2}+\sigma^2_{Y_3}}\right),$$
and for $X_3$ conditioned on the $\mathbf{Y}$ $$P(X_3-Y_1<T)=\Phi\left(\frac{T-\mu_{X_1}-\mu_{Y_1}}{\sigma^2_{X_3}+\sigma^2_{Y_1}}\right),$$ $$P(X_3-Y_2<T)=\Phi\left(\frac{T-\mu_{X_1}-\mu_{Y_2}}{\sigma^2_{X_3}+\sigma^2_{Y_2}}\right),$$ $$P(X_3-Y_3<T)=\Phi\left(\frac{T-\mu_{X_1}-\mu_{Y_3}}{\sigma^2_{X_3}+\sigma^2_{Y_3}}\right).$$
My question is how do I obtain the total probability that $Z$ is less than $T$.
Would it be $P(\mathbf{X}-\mathbf{Y}<T):= \frac{1}{L}\sum^L_{i=1}\sum^L_{j=1}P(X_i-Y_j<T)$?
Appreciate any help and advice.