I know how to find the point of intersection between a line and a plane, but I have no idea how to work backwards and develop a plane equation based on the intersection point, and the vector equation of the line. I was having lots of trouble with these three questions.
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Obviously, for the first part there are many infinitely planes with that condition. In fact, the first question is equal to this question: "How many planes are ther which passing through that point". It is easy to see that we have: $$P_{\alpha}: ax+by+cz=3a-7b-5c$$ where $a,b,c\in\mathbb{R}$. For other parts you just need a plan whose normal vector is perpendicular to the line's leading vector. Indeed, in this case either the line is on the plan or the line is completely outside the plan. Let $\vec{n}=(a,b,c)$ be the normal of that assumed plan. So we should have $-2a+4b+7c=0$. Finding such that constants is not that difficult. For example, $$a=-1, b=-1, c=2/7$$ Now take a point on the line and write the corresponding plane. It is an answer to the 3rd question. And by having a point outside the line you ‘ll have the proper plan answering the 2nd one.
The line we are given is $\ell=\mathbf a+t\mathbf b=(1,-3,2)+t(-2,4,7)$.
For part (a), it's easiest to find the plane perpendicular to $\ell$ and passing through $\mathbf z=(3,-7,-5)$. The normal vector of this plane must be $\mathbf b$. Now find $\mathbf z\cdot\mathbf b$, which works out to be $-69$. Then the equation of the plane is $$\mathbf r\cdot(-2,4,7)=-69$$ For part (c), find a non-zero vector $\mathbf c$ perpendicular to $\mathbf b$ ($\mathbf c\cdot\mathbf b=0$); we'll use $(7,0,2)$. This will be the normal vector of the plane we find. $\mathbf a$ must lie on this plane, so we compute $\mathbf a\cdot\mathbf c=11$. The equation of the plane is thus $$\mathbf r\cdot(7,0,2)=11$$ Part (b) can be answered by changing the scalar in the equation for part (c) to any other value, which will separate the plane and line while keeping them parallel. $$\mathbf r\cdot(7,0,2)=0$$