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For which value on the real constant a were the line L intersects the point (4,1,3), with the direction vector (3,a,2), the plane \[(x,y,z)=(1+t-2s, 2t+s,1+3t+as), t,s \in \mathbb{R}\] in one point.

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it sounds doable, but its hard to make out what its actually asking for
New formulation: For which values of the real constant a does the line L ( L goes trough the point (4,1,3) and has the direction vector (3,a,2)) intersect the plane \[(x,y,z)=(1+t-2s, 2t+s,1+3t+as), t,s \in \mathbb{R}\] in one point.

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Other answers:

we might wanna try to redefine the plane in a format thats easier to parse. (x,y,z)=(1+t−2s,2t+s,1+3t+as),t,s x = (1+t−2s,2t+s,1+3t+as) y = t z = s is there a way to construct that as: a(x-xo)+b(y-yo)+c(z-zo)=0 ??
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 \[t,s \in \]
x = 4 + 3n y = 1 + an z = 3 +2n and x = 1 + t - 2s y = 0 + 2t + s z = 1 + 3t + as (-3 + t - 2s)/3 = n (-1 + 2t + s)/a = n (-2 + 3t + as)/2 = n
(-3 + t - 2s)/3 = n (-1 + 2t + s)/a = (-3 + t - 2s)/3 (-2 + 3t + as)/2 = (-3 + t - 2s)/3 a = (-3 + 6t + 3s)/(-3 + t - 2s) -6 + 9t + 3s(-3 + 6t + 3s)/(-3 + t - 2s) = -6 + 2t - 4s 7t +4s + 3s(-3 + 6t + 3s)/(-3 + t - 2s) = 0 -21t + 7t^2 - 14ts -12s + 4st - 8s^2 + -9s +18st + 9s^2 = 0 -21t + 7t^2 -21s + s^2 +8st = 0 7t^2-21t + s^2 -21s+8st = 0 needs more shuffling, but i believe thats the right idea to take
I think I get the idea, will have to spend some time on i tough, thank you for your help! :)
good luck ;)

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