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Find the equation of the plane that goes through the origin and is parallel to the lines

$$\mathbf{R_1}=3\boldsymbol{\hat{\imath}}+3\boldsymbol{\hat{\jmath}}-\boldsymbol{\hat k} +s(\boldsymbol{\hat \imath}-\boldsymbol{\hat \jmath}-2\boldsymbol{\hat k}),\quad \mathbf{R_2}=4\boldsymbol{\hat\imath}-5\boldsymbol{\hat\jmath}-8\boldsymbol{\hat k}+t(3\boldsymbol{\hat\imath}+7\boldsymbol{\hat\jmath}-6\boldsymbol{\hat k})$$

Additionally, show that one line is on the plane and one is not on the plane

I need some help solving these questions.

So we have two lines that are parallel to the planes and therefore the normal of the plane is perpendicular to the lines

However, How can I find the normal and the plane only with this given information?

Many thanks in advanced

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The normal vector to the plane, which is, as you said, perpendicular to the two lines, you can get by doing to vector product of the two directing vector of each line.

Then you know one point of your plan, so it should be easy to get the equation of the plane.

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