Suppose that the point $\left(\rho,\theta,\phi \right)=\left( 12, \frac{2 \pi}{3}, \frac{3 \pi}{4} \right)$ in spherical coordinates can be

I'm digging deep here, Rekt...!!!....LOL!!!!

First....let's convert these to cylindrical coordinates  (r, θ, z)

r = p sin Φ   =   12 sin 3pi/4  =  6√2

θ = θ = 2pi/3

z = p cos Φ =  12 cos 3pi/ 4 =    -6√2

Then....convert these to rectangular (Cartesian) coordinates (x, y, z)

x = r cos  θ =  6√2 * cos 2pi/ 3 =  6√2 (-1/2) = -3√2

y = r sin θ =   6√2 * sin 2pi/ 3 =  6√2 * √3/2 =  3√6

z =   -6√2

So   (x, y , z)  =   ( -3√2 ,  3√6,  -6√2 )

So.... x + z =  -9√2