The plane that passes through the point and contains the line

Find an equation of the plane. The plane that passes through the point (3, 5, -1) and contains the line x = 4 - t, y = 2t - 1, z = -3t

[@Leonardosanchezandre]

GRACIAS BRO!!!!!!!!!!!!!!!!

[@guggystyle]

You're welcome!

[@pattytheratty]

preciate it dawg 🙏

[@rababrahman9478]

Hi, I was wondering if you gave a detailed explanation on these questions. That would be nice!
Let L1 be the line passing through the point P1=(−5, 2, −6) with direction vector →d1=[0, 1, 1]T, and let L2 be the line passing through the point P2=(7, 0, −18) with direction vector →d2=[4, −3, −3]T.
Find the shortest distance d between these two lines, and find a point Q1 on L1 and a point Q2 on L2 so that d(Q1,Q2) = d. Use the square root symbol '√' where needed to give an exact value for your answer.

[@gntkllrrbt5]

Thanks for the solution!

[@CannedWater_zip]

Thank you this video is totally a lifesaver

[@jiahuizhang8258]

Great explanation

[@TheRedAirOn]

I STRONGLY believe this is all incorrect because vector a is already given which is (3, 5, -1) and the process to find vector a like in this video is wrong.

[@kavishchattoor1729]

P is (3,5,-1) which is a point not a vector which the question clearly stated. Then you have a parametric line which contain Q in the equation. Take those and you get PQ vector. Cross product with the vector of the parametric line, you get the norm.

[@noctao4163]

thank you so much for this

[@mitchellcarl2389]

thank u so much

[@Its.a.me.Gilbert]

thanks

[@WavesGovan335]

감사

[@Raven_Ray1]

Now, how to find the equation if the line and n vector given.