Planes and straight lines

Question

A) Find an equation of the plane $\alpha$ so that it's parallel to the plane $x-2y+\frac{5}{3} z -\sqrt{2}=0$ and contains $(0;0;0)$

B) Given the lines $F: (2x+3y+1=0;y-2z+1=0)$ and $F': (x;y;z)=(1;-1;-2)+t(6;-4;-2) t\epsilon R$ find a plane that contains both of them.

For A), as they are parallel, I can suppose that the normal vector of $\alpha$ is $(1;-2;\frac{5}{3})$ and as it contains $(0;0;0)$ the equation would be $x-2y+\frac{5}{3} z=0$. Is it correct?

For B), firstly I wrote $F$ in parametric form to get its direction vector, which is $(-3;2;1)$. Then, the normal vector of the plane would be the result of the cross product between the direction vector of $F$ and the direction vector of $F'$. That result is $(0;0;0)$, so the plane is 0?

Answer 1

Your solution for part $A$ is correct.

For part $B$ the two lines are parallel so you find the equation of the plane by three non colinear points on the two lines and pass a plane through the points. You will find a valid solution.

Answer 2

Your solution for part A is correct.

For part B, the line $F$ is described as the intersection of two planes. Every plane that contains $F$ has an equation that’s a linear combination of the equations of these two planes $$\lambda(2x+3y+1)+\mu(y-2z+1)=0, \tag1$$ with $\lambda$ and $\mu$ not both zero. If this plane also contains $F'$ then it must contain the point $(1;-1;-2)$, which lies on $F'$, so substitute these coordinates into equation (1) and solve for $\lambda$ and $\mu$. You can eliminate one of these variables by checking one of the two given planes to see if it’s a solution: if it isn’t, then you can set the coefficient of the other plane in (1) to $1$. It turns out that the first plane, $2x+3y+1=0$ does contain $F'$, so you’re done. After obtaining an equation of the plane through $(1;-1;-2)$, you should check that the entire line lies on this plane: generate another point $F'$, say by setting $t=1$, and verify that this point, too, satisfies the plane equation.