1.) Define $P_0 = (0,4,0)$ (can see this is on the plane by setting $x=0$, $y=4$, $z=0$). Let $P_1=(4,-4,3)$. Define $\vec{b} = \vec{P_0P_1} = <4,-8,3>$. Use the normal vector $\vec{n} = <2,-2,5>$. Then the distance from $P$ to the plane is $$D = |comp_{\vec{n}} \vec{b}| = \frac{|\vec{n} \cdot \vec{b}|}{|\vec{n}|} = \frac{39}{\sqrt{33}}.$$

2.) We can reduce this to the same type of problem as number 8.) by picking a point on the first plane and then finding the distance from that point to the other plane. So for example if $x=1$ and $y=z=0$, then the point $P_0 = (1,0,0)$ is on the plane $x-5y-z=1$. Pick $P_1$ on the 2nd plane and define $\vec{b} = \vec{P_0P_1}$. Then use $\vec{n} = <1,-5,-1>$ (or $\vec{n}=<5,-25,-5>$) and compute $|comp_{\vec{n}} \vec{b}|$.

3a.) Set up a system of equations -- its solution is $s=3$, $t=1$.

3b.) Find a point $P_0$ on one of the lines. Take the cross product of both vectors to get a vector perpendicular to both. This is the normal vector of your plane. Use the formula for the equation of a plane given a point on the plane and a normal vector.

4.) a vector equation: $<2,0,1> + t<1,-1,6>$

its parametric equation: $\left\{ \begin{array}{ll} x &= 2+t \\ y &= -t \\ z &= 1 - 6t \end{array} \right.$

its symmetric equation: $x -2 = -y = \frac{z-1}{6}$

5.) Find any point on the line in question -- there's many ways to do this, but perhaps the easiest is to set $z=0$ and solve the resulting system of equations: $$\left\{ \begin{array}{ll} 2x - 5y &= 12 \\ 3x + 4y &= 6. \end{array} \right.$$ The solution will give you an ordered triple that lies on the line. Normal vector to first is $\vec{n}_1 = <2,-5,3>$ and normal vector to second is $\vec{n}_2 = <3,4,-3>$. These vectors are not parallel, so we know the planes intersect. Take their cross product to find a new vector $\vec{v}$ which is perpendicular to both planes (this is because the line whose equation we are trying to find is paralell to both planes, and hence perpendicular to both $\vec{n}_1$ and $\vec{n}_2$!), and therefore parallel to the line of intersection. Use the point you found and this vector $\vec{v}$ to determine the equation of the line.

6.) The parallel vectors are respectively $\vec{v}_1 = <2,4,3>$ and $\vec{v}_2 = <1,3,2>$. Observe they are not parallel because they are not scalar multiples of each other. To see if they are intersecting set up a system of equations, you will find that the system has no solution and therefore the lines do not intersect. Therefore they are skew.

7.) Point on line: $(1,1,1)$, vector $\vec{b} = <1,-1,0>$, vector parallel to line: $\vec{a} = <-2,1,3>$ therefore $$\begin{array}{ll} d &= |\vec{b}| \sin(\theta) \\ &= \frac{|\vec{a} \times \vec{b}|}{|\vec{a}|}. \end{array}$$

8.) Let $P_0 = (0,4,0)$ so that $\vec{P_0 P} = <-1,2,-3>$ then simply compute $$comp_{\vec{n}} \vec{P_0 P} = \frac{|\vec{P_0 P} \cdot \vec{n}|}{|\vec{n}|}.$$