<s>[INST] How can I solve this line & plane intersect question and verify the given answer? Find an equation for the plane that passes through the point $(3,2,1)$ and contains the line of intersection of the planes with equations $x+y+z=3$ and $x+2y+3z=6$.
The given answer from the key is: $-x+2y+5z=6$.
 [/INST] Let  $ p_1 = x+y+z -3$ and let $p_2 = x+2y+3z-6$. So, now, the given plane equations are $p_1 = 0$ and $p_2 = 0$. 
Think about the equation $kp_1 + p_2 = 0$. This is certainly a plane. Also, it contains the intersection of the given two planes, because any point that satisfies both $p_1 = 0$ and $p_2=0$ also satisfies  $kp_1 + p_2 = 0$.
So, by varying $k$, we can produce an entire family of planes that contain the line of intersection. Now we just have to choose the right $k$ to get a specific plane of this family that contains the point $(3,2,1)$.
Now 
\begin{align}
&\;(3,2,1) \text{ lies on the plane } kp_1 + p_2 = 0 \\
\Rightarrow\;& (3,2,1) \text{ satisfies the equation } k(x+y+z -3) + (x+2y+3z-6) =0 \\
\Rightarrow\;& k(3+2+1-3) + (3 +4 +3 -6) = 0  \\
\Rightarrow\;& 3k + 4 = 0  \\
\Rightarrow\;& k = -4/3  \\  
\end{align}
Substitute $k= -4/3$ in $kp_1 + p_2 = 0$. The equation of the desired plane is:
$$
-\tfrac43(x+y+z -3) + (x+2y+3z-6)=0
$$
$$
\text{i.e.  }-4(x+y+z -3) + 3(x+2y+3z-6)=0
$$
$$
\text{i.e.  }-x + 2y +5z = 6 
$$
</s>