Find symmetric equations for the line of intersection of the two planes $x + 2y + 5z = 3$ and $2x + 3y = 1$. The question asks to find symmetric equations for the line of intersection of the two planes $x + 2y + 5z = 3$ and $2x + 3y = 1$. I have done the following work below. Could you please provide feedback on whether or not I have done the question correctly? Thanks!
Since the planes are not scalar multiples of each other, they are not parallel or coincident, so solve the system of equations:
$$
x+2y+5z=3\\
2x+3y=1\\
$$
$$
-2x-4y-5z=-6\\
2x+3y=1\\
$$
$$
-y-5z=-5
$$
$$
\textrm{Let }z=s\\
-y-5s=-5\\
-y=-5+5s\\
y=5-5s, s=\frac{y-5}{5}
$$
$$
2x+3(5-5s)=1\\
2x+15-15s=1\\
2x=-14+15s\\
x=-7+\frac{15}{7}s, s=\frac{x+7}{\frac{15}{7}}=\frac{7x+49}{15}
$$
$$
\therefore\textrm{The symmetric equations for the intersection are:}\\
\frac{7x+49}{15}=\frac{y-5}{5}=z
$$
Edit: I realized the above solution is wrong. I will edit my solution later to my corrected one when I have time.
 A: If you want a more geometry-oriented solution, here we go:
The line of intersection lies on both the planes, so it is perpendicular to both the normals $\vec n_1$ and $\vec n_2$. Thus, the vector $\vec b$ parallel to the line is $\vec n_1\times \vec n_2$.
Now, $$\vec n_1=\hat i+2\hat j+5\hat k \\\, \vec n_2=2\hat i+3\hat j.$$ So $\displaystyle \vec b =-15\hat i+10\hat j-\hat k$. Now all we have to do is find an arbitrary point on the line. Let’s take $z=0$. Then, $x+2y=3$ and $2x+3y=1$ give us the point $(-7,5,0)$  on the line of intersection.
We know, if a point on a line (which is parallel to $\vec b$) has position vector $\vec a$, then the line is $\vec r=\vec a+\lambda\vec b$ where $\lambda$ is a parameter. Thus, any point on the line of intersection has a position vector $$(x,y,z)=(-7-15\lambda, 5+10\lambda, -\lambda)$$ Equating $\lambda$, we get the line $$\frac{x+7}{15}=\frac{5-y}{10}=z$$

Here’s an image to visualise what’s happening.I’ve removed the 3D axes for the sake of clarity.
A: Another way to do it is to find 2 points which should lie on this line;
Let $z=0$
$$x+2y=3\\
2x+3y=1\\x=-7,y=5,z=0 $$
Let $z=1$ $$x+2y=-2\\
2x+3y=1\\x=13,y=-5,z=1$$
So line through these 2 points is:
$$\frac{x+7}{20}=\frac{y-5}{-10}=\frac{z}{1}$$
where $<13,-5,1>-<-7,5,0>=<20,-10,1>$ represents the direction ratios of the line
