How can I solve this line & plane intersect question and verify the given answer?

Question

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: $-x+2y+5z=6$.

Answer 1

There are many ways. Here is one that is straightforward but is not the easiest way.

First, solve the simultaneous equations $x+y+z=3$ and $x+2y+3z=6$. You can do that by substitution or elimination. You will get answers with one parameter: any parameter will do. Then use two different values of that parameter to get two specific solutions to those equations.

You now have three points on your desired plane. You know that your desired plane has the equation

$$ax+by+cz=1$$

for some values of $a$, $b$, and $c$. (There is a possibility that your equation will have a zero on the right-hand side rather than a one, but that does not happen in this problem. I am trying to keep this solution simple.) Substitute your three points, one at a time, into that equation, resulting in three simultaneous equations with the variables $a$, $b$, and $c$. Solve those three simultaneous linear equations in three variables. There are many ways to do that, which you should have learned in Algebra 2, such as substitution, elimination (Gaussian or otherwise), Cramer's rule, and so on.

You now have your $a$, $b$, and $c$, and thus the equation for your desired plane. If $a$, $b$, or $c$ is a fraction, you can multiply the equation by the least common denominator to get all integer constants in your equation.

Answer 2

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: $$ -\tfrac43(x+y+z -3) + (x+2y+3z-6)=0 $$

$$ \text{i.e. }-4(x+y+z -3) + 3(x+2y+3z-6)=0 $$ $$ \text{i.e. }-x + 2y +5z = 6 $$