1. ## The Cross Product

Thanks!!

2. Using the normal, n, that you found and the point P
$n = \left[ {\begin{array}{*{20}c}
3 \\
3 \\
{ - 3} \\
\end{array}} \right]\quad P = \left[ {\begin{array}{*{20}c}
1 \\
0 \\
{ - 2} \\
\end{array}} \right]$

write the equation ot the plane
$\Pi :\;n \cdot (R - P) = 0,\quad R = \left[ {\begin{array}{*{20}c}
x \\
y \\
z \\
\end{array}} \right]$
.

3. ## Re:

I'm still a bit unsure...

Equation of the plane for n(3,3,-3)

3x+3y-3x=0

???

4. It should be written as $3\left( {x - 1} \right) + 3y - 3\left( {z + 3} \right) = 0$.

5. ## Re:

So,

3x-3y-3z=12

Thanks Again,
qbkr21