<s>[INST] 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.
 [/INST] 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
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