1. **State the problem:** We need to evaluate the triple integral $$\int_0^{\pi/2} \int_0^y \int_0^x \cos(x + y + z) \, dz \, dx \, dy.$$\n\n2. **Understand the integral:** The integral is over variables $z$, $x$, and $y$ with limits $z$ from $0$ to $x$, $x$ from $0$ to $y$, and $y$ from $0$ to $\pi/2$. The integrand is $\cos(x + y + z)$.\n\n3. **Integrate with respect to $z$ first:** Treating $x$ and $y$ as constants, we have\n$$\int_0^x \cos(x + y + z) \, dz.$$\nUsing the substitution $u = x + y + z$, $du = dz$, when $z=0$, $u = x + y$, and when $z=x$, $u = x + y + x = y + 2x$.\nSo the integral becomes\n$$\int_{x+y}^{y+2x} \cos(u) \, du = \sin(u) \Big|_{x+y}^{y+2x} = \sin(y + 2x) - \sin(x + y).$$\n\n4. **Now integrate with respect to $x$ from $0$ to $y$:**\n$$\int_0^y [\sin(y + 2x) - \sin(x + y)] \, dx = \int_0^y \sin(y + 2x) \, dx - \int_0^y \sin(x + y) \, dx.$$\n\n5. **Evaluate each integral:**\n- For $\int_0^y \sin(y + 2x) \, dx$, substitute $t = y + 2x$, so $dt = 2 dx$, $dx = \frac{dt}{2}$. When $x=0$, $t=y$, when $x=y$, $t = y + 2y = 3y$.\nThus,\n$$\int_0^y \sin(y + 2x) \, dx = \int_y^{3y} \sin(t) \frac{dt}{2} = \frac{1}{2} [-\cos(t)]_y^{3y} = \frac{1}{2} [-\cos(3y) + \cos(y)].$$\n\n- For $\int_0^y \sin(x + y) \, dx$, substitute $s = x + y$, $ds = dx$. When $x=0$, $s=y$, when $x=y$, $s=2y$.\nSo,\n$$\int_0^y \sin(x + y) \, dx = \int_y^{2y} \sin(s) \, ds = [-\cos(s)]_y^{2y} = -\cos(2y) + \cos(y).$$\n\n6. **Combine the two results:**\n$$\int_0^y [\sin(y + 2x) - \sin(x + y)] \, dx = \frac{1}{2} [-\cos(3y) + \cos(y)] - [-\cos(2y) + \cos(y)] = \frac{1}{2} [-\cos(3y) + \cos(y)] + \cos(2y) - \cos(y).$$\nSimplify:\n$$= -\frac{1}{2} \cos(3y) + \frac{1}{2} \cos(y) + \cos(2y) - \cos(y) = -\frac{1}{2} \cos(3y) + \cos(2y) - \frac{1}{2} \cos(y).$$\n\n7. **Finally, integrate with respect to $y$ from $0$ to $\pi/2$:**\n$$\int_0^{\pi/2} \left(-\frac{1}{2} \cos(3y) + \cos(2y) - \frac{1}{2} \cos(y)\right) dy = -\frac{1}{2} \int_0^{\pi/2} \cos(3y) dy + \int_0^{\pi/2} \cos(2y) dy - \frac{1}{2} \int_0^{\pi/2} \cos(y) dy.$$\n\n8. **Evaluate each integral:**\n- $\int_0^{\pi/2} \cos(3y) dy = \frac{\sin(3y)}{3} \Big|_0^{\pi/2} = \frac{\sin(3\pi/2)}{3} - 0 = \frac{-1}{3} = -\frac{1}{3}.$\n- $\int_0^{\pi/2} \cos(2y) dy = \frac{\sin(2y)}{2} \Big|_0^{\pi/2} = \frac{\sin(\pi)}{2} - 0 = 0.$\n- $\int_0^{\pi/2} \cos(y) dy = \sin(y) \Big|_0^{\pi/2} = 1 - 0 = 1.$\n\n9. **Substitute back:**\n$$-\frac{1}{2} \times \left(-\frac{1}{3}\right) + 0 - \frac{1}{2} \times 1 = \frac{1}{6} - \frac{1}{2} = -\frac{1}{3}.$$\n\n**Final answer:** $$\boxed{-\frac{1}{3}}.$$