For the first one, you want to solve the system $$\begin{cases} x+ 2y +z =4 \\ 2x+y-z=5\end{cases}$$ for $(x,y,z)$. For example, summing I am getting that $x+y=3$. Then I am getting that $z=1-y$, so that $$(x,y,z)=(3-y,y,1-y)=(3,0,1)+y(-1,1,-1)$$ Thus, the intersection is a line $$\mathscr L:(3,0,1)+t(-1,1,-1)\;\;;\; t\in\Bbb R$$
For the second one, suppose we can write your line as $$\mathscr L':w+tv\;\;;\;t\in\Bbb R$$ for some vectors $v,w$. Saying that $\mathscr L'$ is parallel to the line we found above means their directions are parallel. Any line is fully determined by a point and a direction, you should be able to obtain what $\mathscr L'$ is. It seems the exercise wants two lines, one that goes through $A$ and one through $B$.
Recall that if we have a line $v+tw$ and we already know what $w$ is, and know that $v'$ is a point in it, we may let say $t=1$ so that $v+w=v'\implies w=v'-v$, that is, we may let $w$ be $v'$ plus any scalar multiple of the direction we want.
Hint: Any plane passing through intersection of $4x - 2y + z - 3 = 0$ and $2x - y + 3z + 1 = 0$ is given by $$(4x - 2y + z - 3) + k(2x - y + 3z + 1) = 0$$ or what is the same as $$(2k + 4)x - (k + 2)y + (3k + 1)z + k - 3 = 0$$
This is perpendicular to $3x + y - z + 7 = 0$. Using dot product of normal vectors you can now find $k$.
EDIT: If you do calculations you will find $k = -9/2$ and final answer would be same as that provided in another answer namely $-2x + y - 5z = 3$ or $2x - y + 5z + 3 = 0$
Best Answer
Use the cross product of the normals of the two given planes, and the given point that the third plane passes through, to form the dot-product vector equation of the third plane.