Remember the general equation of a line in vector form:

$\displaystyle \vec{r}=r_0+t\vec{v}=\left \langle x_0,y_0,z_0\right \rangle+t\left \langle a,b,c\right \rangle$, where $\displaystyle \left \langle x_0,y_0,z_0\right \rangle$ is the starting point, and $\displaystyle \left \langle a,b,c\right \rangle$ is the difference between the start and ending points.

Lets apply this to our problem.

$\displaystyle \vec{r}=\left \langle 2, 10, -6\right \rangle+t\left \langle 5, 6, -20\right \rangle$

Distribute the $\displaystyle t$

$\displaystyle \vec{r}=\left \langle 2, 10, -6\right \rangle+\left \langle 5t, 6t, -20t\right \rangle$

Now we simply do vector addition to get

$\displaystyle \vec{r}=\left \langle 2+5t, 10+6t, -6-20t\right \rangle$

Finding the angle between two planes requires us to find the angle between their normal vectors.

To obtain normal vectors, we simply take the coefficients in front of $\displaystyle x,y,z$.

$\displaystyle \mathbf{n_1} = < 4,-3,-2>$

$\displaystyle \mathbf{n_2} = < 12,2,-7>$

The (acute) angle between any two vector is

$\displaystyle \theta = \cos^{-1}(\frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|\|\mathbf{b}\|})$,

Substituting, we have

$\displaystyle \theta = \cos^{-1}(\frac{< 4,-3,-2> \cdot < 12,2,-7>}{\sqrt{16+9+4}\sqrt{144+4+49}})$

$\displaystyle \theta = \cos^{-1}(\frac{< 4,-3,-2> \cdot < 12,2,-7>}{\sqrt{16+9+4}\sqrt{144+4+49}})$

$\displaystyle \theta \approx \cos^{-1}(\frac{56}{75.584})$

$\displaystyle \theta \approx 42.192$.

Substituting the components of the line into those of the plane, we have

$\displaystyle 2(2t+4)+(t-1)+(-t)=9$

$\displaystyle 4t+8+t-1-t=9$

$\displaystyle t = \frac{1}{2}$

Substituting this value of $\displaystyle t$ back into the components of the line gives us

$\displaystyle (5,-\frac{1}{2},-\frac{1}{2})$.

A quick way to notice the answer is $\displaystyle 0$ is to notice the planes are parallel (They only differ by the constant on the right side).

Typically though, to find the angle between two planes, we find the angle between their normal vectors.

A vector normal to the first plane is $\displaystyle \mathbf{n}_1 = < 1,1,1>$

A vector normal to the second plane is $\displaystyle \mathbf{n}_2 = < 1,1,1>$

Then using the formula for the angle between vectors, $\displaystyle \theta = \cos^{-1} (\frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|})$, we have

$\displaystyle \theta = \cos^{-1} (\frac{< 1,1,1> \cdot < 1,1,1>}{\sqrt{1^2+1^2+1^2} \sqrt{1^2+1^2+1^2}}) = \cos^{-1}(1) = 0$.

The equation of a plane is defined as

$\displaystyle n\bullet< x-x_0,y-y_0,z-z_0>=0$

where $\displaystyle n$ is the normal vector of the plane. 

To find the normal vector, we first get two vectors on the plane 

$\displaystyle OP=< 5,4,0>$ and $\displaystyle OQ=< -3,-7,0>$

and find their cross product. 

The cross product is defined as the determinant of the matrix

$\displaystyle \begin{vmatrix} i&j &k \\ 5&4 &0 \\ -3&-7 &0 \end{vmatrix}$

Which is

$\displaystyle =det(\begin{vmatrix} 4&0 \\ -7&0 \end{vmatrix})i-det(\begin{vmatrix} 5&0 \\ -3&0 \end{vmatrix})j+det(\begin{vmatrix} 5&4 \\ -3&-7 \end{vmatrix})k$

$\displaystyle =(4*0-(-7)*0)i-(5*0-(-3)*0)j+(5*(-7)-(-3)*4)k$

$\displaystyle =-23k$

Which tells us the normal vector is 

$\displaystyle < 0,0,-23>$

Using the point $\displaystyle O$ and the normal vector to find the equation of the plane yields

$\displaystyle < 0,0,-23>\bullet < x-0,y-0,z-0>=0$

$\displaystyle < 0,0,-23>\bullet < x,y,z>=0$

$\displaystyle 0x+0y+(-23)z=0$

Simplified gives the equation of the plane 

$\displaystyle z=0$

The equation of a plane is defined as

$\displaystyle n\bullet< x-x_0,y-y_0,z-z_0>=0$

where $\displaystyle n$ is the normal vector of the plane. 

To find the normal vector, we first get two vectors on the plane 

$\displaystyle OP=< 0,1,5>$ and $\displaystyle OQ=< 0,-2,-4>$

and find their cross product. 

The cross product is defined as the determinant of the matrix

$\displaystyle \begin{vmatrix} i&j &k \\ 0&1 &5 \\ 0&-2 &-4 \end{vmatrix}$

Which is

$\displaystyle =det(\begin{vmatrix}1&5 \\ -2&-4 \end{vmatrix})i-det(\begin{vmatrix} 0&5 \\ 0&-4 \end{vmatrix})j+det(\begin{vmatrix} 0&1 \\ 0&-2 \end{vmatrix})k$

$\displaystyle =(1*(-4)-(-2)*5)i-(0*(-4)-(0)*5)j+(0*(-2)-0*1)k$

$\displaystyle =6i$

Which tells us the normal vector is 

$\displaystyle < 6,0,0>$

Using the point $\displaystyle O$ and the normal vector to find the equation of the plane yields

$\displaystyle < 6,0,0>\bullet < x-0,y-0,z-0>=0$

$\displaystyle < 6,0,0>\bullet < x,y,z>=0$

$\displaystyle 6x+0y+0z=0$

Simplified gives the equation of the plane 

$\displaystyle x=0$

Planes that are parallel to each other only differ (if at all) by the constant on the right-hand side (when both sides are simplified). Since $\displaystyle x+4y-3z =5$ has the same $\displaystyle x,y,z$ coefficients as the given plane, they are parallel to each other.

We can begin by rewriting the expression for the cylinder as follows

$\displaystyle 9x^2 + y^2 = 9 \Rightarrow x^2 + \left( \frac{y}{3} \right) ^2 = 1$.

This tells us that $\displaystyle x = \cos t, y = 3 \sin t$.  Plugging this back into the equation for the plane $\displaystyle x + y + z = 7$ to find $\displaystyle z = 7 - x - y = 7 - \cos t - 3 \sin t$.

This gives us the representation of the curve of intersection as

$\displaystyle \vec{r}(t)= \left< \cos t, 3 \sin t, 7-\cos t - 3 \sin t\right>$.

The equation of a plane is defined as

$\displaystyle n\bullet< x-x_0,y-y_0,z-z_0>=0$ 

where 

$\displaystyle n$ 

 is the normal vector of the plane. 

To find the normal vector, we first get two vectors on the plane 

 $\displaystyle OP=< 1-2,5-3,3-4>=< -1,-2,-1>$

 and 

 $\displaystyle OQ=< 5-2,2-3,1-4>=< 3,-1,-3>$

and find their cross product. 

The cross product is defined as the determinant of the matrix

$\displaystyle \begin{vmatrix} i&j &k \\ -1&-2 &-1 \\ 3&-1 &-3 \end{vmatrix}$ 

Which is

 $\displaystyle =det(\begin{vmatrix} -2&-1 \\ -1&-3 \end{vmatrix})i-det(\begin{vmatrix} -1&-1 \\ 3&-3 \end{vmatrix})j+det(\begin{vmatrix} -1&-2 \\ 3&-1 \end{vmatrix})k$

$\displaystyle =(-2(-3)-(-1)*(-1))i-(-1(-3)-3(-1))j+(-1(-1)-3(-2))k$ 

$\displaystyle =5i-6j+7k$ 

Which tells us the normal vector is 

 $\displaystyle < 5,-6,7>$

Using the point 

 $\displaystyle O=(2,3,4)$

 and the normal vector to find the equation of the plane yields

 $\displaystyle < 5,-6,7>\bullet< x-2,y-3,z-4>=0$

$\displaystyle 5(x-2)+(-6)(y-3)+7(z-4)=0$ 

 $\displaystyle 5x-10-6y+18+7z-28=0$

Simplified gives the equation of the plane 

$\displaystyle 5x-6y+7z=20$

To find the angle between the planes, we find the angle between their normal vectors.

We have

$\displaystyle \mathbf{n}_1 = < 1,5,-1>$, for the first plane, and

$\displaystyle \mathbf{n}_2 = < 1,0,-1>$, for the second plane.

The angle between these two vectors is

$\displaystyle \theta = \cos^{-1}(\frac{\mathbf{n}_1 \cdot \mathbf{n}_2}{\|\mathbf{n}_1\|\|\mathbf{n}_2\|})$

$\displaystyle = \cos^{-1}(\frac{< 1,5,-1> \cdot < 1,0,-1>}{\sqrt{1^2+(-5)^2+(-1)^2}\sqrt{1^2+0^2+(-1)^2}})$

$\displaystyle \approx \cos^{-1}(\frac{2}{7.34847}) \approx 74.207$