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it sounds doable, but its hard to make out what its actually asking for

the line equation parts are
x = 4 + 3n
y = 1 + an
z = 3 +2n

I think it's x=1+t-2s, y= 2t+s, z=1+3t+as and the t,s is just to say that thet are Real

hmm, then we might be able to use x and y to solve for n, and use n to solve for a ... is my idea

\[(x,y,z)=(1+t-2s),2t+s, 1+3t+as)\]
\[x=1+t-2s\]
\[y=2t+s\]
\[z=1+3t+as\]
where \[t,s \in \]

I think I get the idea, will have to spend some time on i tough, thank you for your help! :)

good luck ;)