Why is sum of squares equal to eigenvalue in PCA? We fit a line or a hyperplane to a set of points. We project the points onto the hyperplane. The sum of squared distances of the projected points to origin is equal to the eigenvalue. Why is that?

Let's presume, without loss of generality, that the data is centered, i.e. the mean is equal to zero.
First point:
Since the dataset $X = \{\mathbf{x}_i\}_{i=1}^n$ is centered, we have:
$$
n\sigma_k^2 = n \cdot \frac{1}{n}\sum_{i=1}^n x_{ik}^2 = \sum_{i=1}^n x_{ik}^2,
$$
where $x_{ik}$ is the $k$-th coefficient of the $i$-th data point. Now consider some unit vector $\mathbf{v}$. The sum of squared distances of the data points in the direction of $\mathbf{v}$ is:
$$
\sum_{i=1}^n \langle \mathbf{v}, \mathbf{x}_i \rangle^2 = \sum_{i=1}^n \sum_{k=1}^d v_k^2 x_{ik}^2 = \sum_{k=1}^d v_k^2 \sum_{i=1}^n x_{ik}^2 = \sum_{k=1}^d v_k^2 n\sigma_k^2 = n\langle \mathbf{v}, \Sigma \mathbf{v} \rangle,
$$
where $\langle \cdot, \cdot \rangle$ is the inner product.
Furthermore, for an $m$-dimensional linear subspace $V$ spanned by the orthonormal vectors $(\mathbf{v}_1, \ldots, \mathbf{v}_m)$, the squared distance of the projection of some $\mathbf{x}_i$ to this subspace is $\sum_{r=1}^m \langle\mathbf{v}_r, \mathbf{x}_i\rangle^2$, so we can generally state: The sum of squared norms of the projections of the dataset $X$ to some subspace $V$, denoted by $SSP_V(X)$, is:
$$
SSP_V(X) = \sum_{i=1}^n \sum_{r=1}^m \langle\mathbf{v}_r, \mathbf{x}_i\rangle^2 = n\sum_{r=1}^m \langle\mathbf{v}_r, \Sigma \mathbf{v}_r\rangle.
$$
Second point:
PCA finds the $V$ that maximizes $SSP_V(X)$. But this is a quadratic optimization problem with quadratic constraints (because the $\mathbf{v_i}$ are unit vectors), which can be solved e.g. by using Lagrange multipliers. The result is, that the linear subspace $V$ is the one spanned by the eigenvectors $(\omega_r)_{r=1}^m$ of $\Sigma$ that belong to the $m$ largest eigenvalues. And, if you enter an orthonormal selection of those into the expression of $SSP_V(X)$ above, you obtain:
$$
\begin{align}
SSP_V(X) &= n\sum_{r=1}^m \langle\mathbf{\omega}_r, \Sigma \mathbf{\omega}_r\rangle\\
         &= n\sum_{r=1}^m \langle\mathbf{\omega}_r, \lambda_r \mathbf{\omega}_r\rangle\\
         &= n\sum_{r=1}^m \lambda_r.
\end{align}
$$
Thus, the sum of squared distances of the projected points to the origin is equal to the size $n$ of the dataset $X$ times the sum of the first $m$ eigenvalues of $\Sigma$.
