I know the following results:
$\frac{RSS}{\sigma^{2}} \sim \chi^{2}_{n-r}$ where $r = rank(X)$ and $k$ is the degrees of freedom. Then, for any $c \in \mathbb{R^{p}} $, and when X has full rank, i.e. $rank(X) = p$, $\frac{\mathbf{c}^{T}\hat{\beta}-\mathbf{c}^{T}\beta}{\sqrt{\mathbf{c^{T}(X^{T}X)^{-1}\mathbf{c}\frac{RSS}{n-p}}}} \sim t_{n-p}$, following from $\frac{\mathbf{c^{T}}\hat{\beta}-\mathbf{c^{T}}\beta}{\mathbf{c^{T}}(X^{T}X)^{-1}\mathbf{c}\sigma^{2}} \sim N(0,1)$.
Suppose that we fit a linear model with design matrix X to obtain estimates $\boldsymbol{\hat{\beta}}$ and $\hat{\sigma}^{2}$. For given covariates, $x_{*}$, the predicted response is $\hat{y}_{*} = x_{*}^{T}\boldsymbol{\hat{\beta}}$.
We also know that $Var(x_{*}^{T}\hat{\beta}) = x_{*}^{T}(X^{T}X)^{-1}x_{*}\sigma^{2}$.
How, from the above, can I get the following result? A $100(1-\alpha)$ confidence interval for a single future response is $\hat{y_{*}} \pm t_{n-p}^{(\frac{\alpha}{2})}\hat{\sigma}\sqrt{1+x_{*}^{T}(X^{T}X)^{-1}x_{*}}$? I've been stuck for ages.