Given the line $(x, y, z) = (1, -3, 2) + t(-2, 4, 7)$, find planes to satisfy the following conditions:
A plane which is intersected by the line at the point $(3, -7, -5)$,
A plane which is not intersected by the line,
A plane that the line lies on.
The course content doesn't cover this specifically so I have no clue how to approach answering it.
I'll get you started and then you try to finish it. This will give a plane NOT intersected by the line.
To find a vector that is perpendicular to your line, form the dot product of the line's direction vector and this unknown orthogonal vector. Like this:
$$\left(\begin{array}{c}-2\\4\\7\end{array}\right)\cdot\left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{c}0\\0\\0\end{array}\right)$$
Next, pick two values for $x$ and $y$. It doesn't matter, but I like to pick $x=4$ and $y=2$. (reverse x & y and change the sign of one of them). $$\left(\begin{array}{c}-2\\4\\7\end{array}\right)\cdot\left(\begin{array}{c}4\\2\\z\end{array}\right)=\left(\begin{array}{c}0\\0\\0\end{array}\right)$$ Now solve for z. You can pretty much see by inspection that z will be 0. Thus an orthogonal vector to the line is (4,2,0). Now pick a point that is not on the line. Let's call it P and you figure out what coordinates it has. A plane that is parallel to the line and passes through point P will be $$\left(\begin{array}{c}4\\2\\0\end{array}\right)\cdot\left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{c}4\\2\\0\end{array}\right)\cdot\left(\begin{array}{c}P_{x}\\P_{y}\\P_{z}\end{array}\right)$$ By & by, it is totally possible to pick a P that is not on your line, but still causes this plane to pass through your line, so be careful.