From the second equation, x = y-z+1.
Substitute for x in the other equation: (y-z+1)+2y+3z=1
y = (-2/3)z
x = y-z+1
Let z = t, then parametric equations for the line are: x = (-5/3)t+1
y = (2/3)t
z = t
To get the symmetric equations, solve each parametric equation for t:
t = (x-1)/(-5/3) t = (y-0)/(2/3) and t = (z-0)/1
Symmetric equations: (x-1)/(-5/3) = (y-0)/(2/3) = (z-0)/1
Note that (1,0,0) is a point on the line of intersection (obtained by setting t=0 in the parametric equations) and the vector
<-5/3, 2/3, 1> is parallel to the line of intersection (-5/3,2/3,and 1 are the coefficients of t in the parametric equations).