3D: 2 planes intersecting

I believe I've figured out a and b, but I don't understand c

Consider the points A(1,2,0), B(1,0,1), C(0,1,-1) and D(1,0,-1)

a) Find the equation of the plane passing through the points A, B, C. Is the point D on this plane?

b)Find a parametric equation for the line passing through D which is perpendicular (the line) to the plane of the triangle ABC.

c) Find the equation of the plane passing through A and D, which is perpendicular to the plane of the triangle ABC. Note: Two planes are perpendicular if their normal vectors are perpendicular.

a)equation of the plane ABC is -3x+y+2z=-1 and D is not on the plane

$\displaystyle \begin{gathered}

\overrightarrow u = < 0,2, - 1 > \hfill \\

\overrightarrow v = < - 1,1, - 2 > \hfill \\

\end{gathered}

$

$\displaystyle \begin{gathered}

\overrightarrow n = - 3i + j + 2k \hfill \\

\overrightarrow r - \overrightarrow {{r_o}} = < x - 0,y - 1,z - ( - 1) > \hfill \\

\end{gathered}

$

$\displaystyle \begin{gathered}

\overrightarrow r - \overrightarrow {{r_o}} \overrightarrow n = 0 \hfill \\

< x - 0,y - 1,z - ( - 1) > < - 3,1,2 > = 0 \hfill \\

\end{gathered}

$

$\displaystyle \begin{gathered}

- 3x + 3 + y + 2z - 2 = 0 \hfill \\

- 3x + y + 2x = - 1 \hfill \\

\end{gathered}

$

b)equation of line perpendicular to plane:

$\displaystyle \begin{gathered}

\overrightarrow {{r_o}} = < 1,0, - 1 > \hfill \\

\overrightarrow v = < - 1,1, - 2 > \hfill \\

\end{gathered}

$

$\displaystyle \begin{gathered}

\overrightarrow r = \overrightarrow {{r_o}} + t\overrightarrow v \hfill \\

< x,y,z > = < 1,0, - 1 > + t < - 1,1, - 2 > \hfill \\

\end{gathered}

$

x=1-t

y=t

z=-1-2t

and like I said, I have no idea what to do for c