, 2021

Solution:
Introduce coordinates so that $\mathcal{X}, \mathcal{Y}, \mathcal{Z}$ are normal to $(1,0,0), (0,1,0)$, and $(0,0,1)$, respectively. Also, suppose that $\mathcal{P}$ is normal to unit vector $(\alpha, \beta, \gamma)$ with $\alpha, \beta, \gamma \geq 0$.

Since the area of $\mathcal{X}$ is $1$, the area of its projection is the absolute value of the cosine of the angle between $\mathcal{X}$ and $\mathcal{P}$, which is $|(1,0,0) \cdot (\alpha, \beta, \gamma)| = \alpha$. (For parallelograms it suffices to use trigonometry, but this is also true for any shape projected onto a plane. One way to see this is to split the shape into small parallelograms.) Similarly, $y = \beta$ and $z = \gamma$. Therefore $x^2 + y^2 + z^2 = 1$, from which it is not hard to calculate that $(x, y, z) = (6/19, 10/19, 15/19)$. Therefore $x + y + z = 31/19$.