Answer
$(x,y,z)=(\dfrac{6 \sqrt 2}{2},\dfrac{-6\sqrt 2}{2}, 2)$
Work Step by Step
Relationship between cylindrical co-ordinates and rectangular co-ordinates is: $$x= r \cos \theta ; y= r \sin \theta; z=z ~~~(1)$$
We are given that $r=6$ and $\theta=\dfrac{-\pi}{4}$
Plug these values in Equation (1) to obtain: $$x=6 \cos (\dfrac{-\pi}{4})=(6) \cos (\dfrac{\sqrt 2}{2})=\dfrac{6 \sqrt 2}{2} \\ y= 6 \sin (\dfrac{-\pi}{4})=-(6) \sin (\dfrac{\sqrt 2}{2})=-\dfrac{6 \sqrt 2}{2} \\z= 2$$
So, our solution is: $(x,y,z)=(\dfrac{6 \sqrt 2}{2},\dfrac{-6\sqrt 2}{2}, 2)$
