1.) Using the formula for the equation of a line through a point parallel to the normal vector $<1,2,3>$, we get $$\vec{r} = <2,1,-3> + t<1,2,3> = <2+t,1+2t,-3+3t>.$$

2.) Since the line perpendicular to the plane is parallel to the normal vector $\vec{n} = <3,-2,4>$, we see that the line in question is $$\vec{r} = <-1,3,2> + t<3,-2,4> = <-1+3t,3-2t,2+4t>.$$

3.) First form the vector $\vec{PQ} = <-3,1,-2>$. Then use the formula for the line parallel to $\vec{PQ}$ through the point $(1,2,3)$ (or the point $(-2,3,1)$, it doesn't matter) as we did in problems 1 and 2 above.

4.) They are not parallel because $L_1$'s corresponding vector $<2,3,-1>$ and $L_2$'s corresponding vector $<1,1,3>$ are not scalar multiples of each other. To see if they intersect, consider the system of equations $$\left\{ \begin{array}{ll} 1+2t & -1+s \\ 3t &= 4+s \\ 2-t &= 1 + 3s \end{array} \right.$$ This system has no solution, so we see that the lines do not intersect. Therefore the answer is neither.

5.) Directly from the scalar equation of the plane through $(1,2,3)$ with normal vector $<3,-5,1>$, we see the equation of the plane is $$3(x-1)-5(y-2)+(z-3)=0.$$

6.) It intersects when $t = \frac{3}{2}$, so $x=4, y=2,$ and $z=-\frac{7}{2}$. So the point of intersection is $(4, -\frac{5}{2}, -\frac{7}{2})$.

7.) Neither. The normal vector for the 1st is $<1,1,1>$ and the normal vector for the 2nd is $<1,-1,1>$. So $<1,1,1> \cdot <1,-1,1> = 1$ so we can conclude they are not perpendicular. Also $<1,1,1>$ is not a scalar multiple of $<1,-1,1>$, so they are not parallel.

8.) $proj_{\vec{y}} \vec{x} = \frac{\vec{y} \cdot \vec{x}}{|\vec{y}|^2} \vec{y} = \frac{2}{2} \vec{y} = \vec{y}$

9.) They are orthogonal because of the box on page 817 and the following computation: $$<0,1> \cdot <1,0> = 0.$$