Show that planes $x+2y+3z=8$ and $2x+3y+4z=11$ intersect in a line coplanar with $\frac{x+1}{1}=\frac{y+1}{2}=\frac{z+1}{3}$ Question is

Show that the line of intersection of the planes
  $$ x + 2y + 3z = 8 \quad\text{and}\quad 2x + 3y + 4z = 11 $$
  is coplanar with the line
  $$\frac{x+1}{1}=\frac{y+1}{2}=\frac{z+1}{3}$$
Also find the equation of the plane containing them. 

$\text{Any hint how could I proceed ?}$
I know how to find the vector the line of intersection would be parallel to , which is given by cross product of the normal vectors of the two planes .
But that does not help in finding the equation of the line of intersection .
What I know is that to prove that two lines are coplanar we have to show that they intersect i.e the shortest distance between them is zero .
But unfortunately this formula again requires one point through which the line of intersection would pass which is unknown.
ps - I am in school yet :)

The line of intersection between the two planes can be represented parametrically. Let $z = t$:
$$
\left\{ 
\begin{array}{c}
x + 2y = 8 - 3t \\  
2x + 3y = 11 - 4t  \\ 
\end{array}
\right. 
$$
Then $2 (1) - (2)$ gives $y = (16 - 6t) - (11 - 4t) = 5 - 2t$, and so $x = (8 - 3t) - 2(5 - 2t) = -2 + t$.
We have to check that these lines are not skew. This is a sufficient condition for two lines in 3D space to be coplanar, as two lines can at most span $\mathbb R^2$ (imagine rotating a plane around one line until it intersects the other line).
The other line can be rewritten parametrically as $(\lambda - 1, 2 \lambda - 1, 3 \lambda - 1)$. Thus we have:
$$
\left\{ 
\begin{array}{c}
t - 2 = \lambda - 1 \\  
-2t + 5 = 2 \lambda - 1 \\ 
t = 3 \lambda - 1 \\ 
\end{array}
\right. 
$$
Substituting the third into the first, we have $(3 \lambda - 1) - 2 = \lambda - 1 \implies 3 \lambda - 3 = \lambda - 1$, thus $\lambda = 1$ and $t = 2$. Now we have to verify that $-2 \cdot 2 + 5 = 2 \cdot1 - 1$. This is true, so the lines are coplanar.
The equation of the plane is given by the normal vector to the direction vectors of the lines. This means we have to compute $(1, -2, 1)^T \times (1, 2, 3)^T$, which is $(-6-2, 1-3, 2--2)^T$ or $(-8, -2, 4)^T$. A multiple of this normal vector is $(4, 1, -2)^T$, so we need:
$$\begin{pmatrix} 4 \\ 1 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} $$
or $4x + y - 2z = -3$.
