= Solution
Write $\mathbf1$ for the all-ones vector and set
$$
P_0=\frac{\mathbf1\mathbf1^{\mathsf T}}n,\qquad
P_X=X(X^{\mathsf T}X)^{-1}X^{\mathsf T},\qquad
\nu=n-p-1.
$$
The centering assumption makes $P_0P_X=0$. Thus the full <normal linear model> has fitted-value <orthogonal projection> $P_0+P_X$, and its <residual sum of squares> is $R=Y^{\mathsf T}(I-P_0-P_X)Y$. Assume $\nu>0$, as is needed to estimate the unknown error <variance> from residuals.
For the null that all slopes vanish, the reduced fitted value is $\bar Y\mathbf1$, with residual sum of squares $R_0=Y^{\mathsf T}(I-P_0)Y$. The <F-test> is
$$
\boxed{F=\frac{(R_0-R)/p}{R/\nu}
=\frac{Y^{\mathsf T}P_XY/p}{R/\nu}\sim F_{p,\nu}\quad\text{under the null}}.
$$
Reject for an upper-tail value exceeding the $1-\alpha$ quantile of this <F-distribution>. The required theorem is <Cochran's theorem>: for an isotropic <normal distribution>, projections onto orthogonal subspaces are <independent>, and the squared norm on a rank-$r$ subspace divided by $\sigma^2$ has a <chi-squared distribution> with $r$ degrees of freedom. Here the null makes the projected mean in $\operatorname{col}(X)$ zero, so $(R_0-R)/\sigma^2\sim\chi_p^2$, independently of $R/\sigma^2\sim\chi_\nu^2$.
Let $p_1,p_2$ be the numbers of columns in the two predictor blocks. To test the first block while retaining the second as nuisance, form
$$
P_2=X_2(X_2^{\mathsf T}X_2)^{-1}X_2^{\mathsf T},\qquad
R_2=Y^{\mathsf T}(I-P_0-P_2)Y.
$$
The <partial F-test for nested linear models> is
$$
\boxed{F_1=\frac{(R_2-R)/p_1}{R/\nu}\sim F_{p_1,\nu}\quad\text{when }\beta_1=0}.
$$
Again reject in the upper tail. The difference of the nested fitted-value projections has rank $p_1$, is orthogonal to the full residual projection, and annihilates the null mean, which gives the same <independent> chi-squared argument. If a block has no columns, the corresponding test is vacuous.
The phrase <orthogonal statistical parameters> refers to zero cross-block <Fisher information>, not to the numerical parameter vectors being perpendicular. In this model the cross-block information is $X_1^{\mathsf T}X_2/\sigma^2$, so the <orthogonal coefficient blocks in a centered normal linear model> condition is
$$
\boxed{X_1^{\mathsf T}X_2=0}.
$$
Then the two fitted subspaces are orthogonal, the block <least-squares estimators> have zero <covariance> and are <independent> because they are jointly Gaussian, and each block's <extra sum of squares> is unchanged by including the other block. Centering also makes both blocks orthogonal to the intercept.
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