You have here two of the fundamental ways to represent a line in $\mathbb R^2$. The first is described by a parametric representation that uses a point $\mathbf p_0$ on the line and a direction vector $\mathbf v$ parallel to the line. Every point on the line can then be expressed in the form $\mathbf p_0+\lambda\mathbf v$, $\lambda\in\mathbb R$.
The second line is described implicitly by an equation in point-normal form. There are various ways to interpret it geometrically, but I find this one easiest to visualize: The perpendicular line through the origin to a line $\mathscr l$ is completely characterized by a direction vector $\mathbf n$. Per the preceding paragraph, a parameterization of that perpendicular line is simply $\lambda\mathbf n$. If $\mathscr l$ is translated so that it passes through the origin, the translated line $\mathscr l'$ has the property that the position vector $\mathbf x'$ of every point on it is orthogonal to $\mathbf n$. That is, $\mathscr l'$ is the set of points that satisfy $\mathbf n\cdot\mathbf x'=0$. We can translate $\mathscr l$ in this way by fixing some point $\mathbf p$ on it and subtracting this from every point $\mathbf x$ on $\mathscr l$. Substituting into the orthogonality condition, an equation for $\mathscr l$ is therefore $\mathbf n\cdot(\mathbf x-\mathbf p)=0$, or $\mathbf n\cdot\mathbf x=\mathbf n\cdot\mathbf p$. Observe that it doesn’t matter which point was chosen for $\mathbf p$—the last form of the point-normal equation says that the dot product of any point on $\mathscr l$ with the normal vector $\mathbf n$ is constant.
Multiplying both sides of that last equation by $\mathbf n/\|\mathbf n\|^2$, we have $${\mathbf n \cdot \mathbf x\over \mathbf n\cdot\mathbf n}\mathbf n = {\mathbf n \cdot \mathbf p\over \mathbf n\cdot\mathbf n}\mathbf n,$$ which says that every point of $\mathscr l$ has the same orthogonal projection onto the normal vector $\mathbf n$. The length of this projection is $|\mathbf n\cdot\mathbf p|/\|\mathbf n\|$, which is also the distance of the line from the origin. Thus, the constant term in the point-normal equation of a line is the distance to the origin multiplied by the length of the normal vector. The sign of the constant term tells you on which side of the origin relative to the direction $\mathbf n$ the line lies.
To convert from the point-direction form to point-normal form, you have to find a normal to the line, which is any vector perpendicular to the line’s direction vector, and vice-versa when converting from point-normal to point-direction. The constant term for the point-normal form is then found by computing the dot product of this normal vector with any point on the line, as described above.
It’s not really necessary to convert representations to solve this problem, though. The intersection of the two lines is a point on the first line that satisfies the equation of the second, so you could simply plug the generic expressions for $x$ and $y$ from the first into the second and solve for $\lambda$ without doing any conversion. I expect that your professor has you do the conversion because he wants you to practice solving systems of linear equations, which is a major motivation for linear algebra in the first place.
You did right the first step: if the planes have a common point the line shall pass through it.
However the planes do not need in general to have a common point.
The concept to apply is that, given two planes, the points equi-distant from them lie on one of the two planes bisecting the angles between the given planes. If these are parallel, then there is only one bi-secting plane (that in between) if they are distinct, or infinite if they are coincident .
Then, given two non-parallel planes $\pi_1=0, \quad \pi_2=0$ with unitary normal vectors $\bf n_1,\; \bf n_2$, the bisecting plane :
- belongs to the sheaf $\lambda \pi_1+ \mu \pi_2=0$
- has a unitary normal vector proportional to $\bf n_1+\bf n_2$ (external angle) or $\bf n_1-\bf n_2$ (internal angle).
Thus, having three planes,
- take two couples of them (e.g. $\pi_1,\,\pi_2$ and $\pi_2,\,\pi_3$)
- determine the four (or less, if you do not use homogeneous coordinates) bisecting planes $\pi_{1,2,a},\, \pi_{1,2,b}, \, \pi_{2,3,a},\, \pi_{2,3,b}$
- any line given by the crossing of two planes $\pi_{1,2,x}$ & $\pi_{2,3,y}$ will have $d_1=d_2\,\& \,d_2=d_3$.
In conclusion, for three non-parallel planes we have $4$ equi-distant lines. Less than that if some of the planes are parallel.
The above when the distance is measured in absolute terms. If on each plane
a direction of its normal is chosen as to measure the distance in algebraic ($\pm$) terms,
then the line is unique (or does not exist).
To better visualize the whole situation, let's reduce the problem in 2D.
Given three non-parallel lines, thus a non-degenerate triangle made by them,
the points that have the same absolute distance from the three lines are $4$:
the $C_k$ shown in the sketch.
Coming to your particular case, the three planes are concurrent
in one point: the system has only one solution $P=(1,-1,-1)$.
The unit normals to the planes are
$$
\eqalign{
& {\bf n}_{\,1} = 1/3\left( {1,2,2} \right) \cr
& {\bf n}_{\,2} = 1/3\left( {1, - 2,2} \right) \cr
& {\bf n}_{\,3} = 1/3\left( {2,1,2} \right) \cr}
$$
Four bisecting planes are
$$
\eqalign{
& \pi _{1,2,a} = x + 0y + 2z + 1 = x + 2z + 1 = 0 \cr
& \pi _{1,2,b} = 0x + 2y + 0z + 2 = y + 1 = 0 \cr
& \pi _{2,3,a} = {3 \over 2}x - {1 \over 2}y + 2z + 0 = 3x - y + 4z = 0 \cr
& \pi _{2,3,b} = - {1 \over 2}x - {3 \over 2}y + 0z - 1 = x + 3y + 2 = 0 \cr}
$$
they are of course all passing through the point P.
Then starting and taking $\pi _{1,2,a} $ and $ \pi _{2,3,a} $, the cross product of their normals is $(2,2,-1)$.
Therefore a first line is
$$
l_{\,1} :\;{{x - 1} \over { 2}} = {{y + 1} \over 2} = {{z + 1} \over { - 1}} = t
$$
In fact, inserting its generic point $P_1(t)=(1+2t,-1+2t,-1-t)$ into the (normalized) equations of the three planes
we get $4/3t(1,-1,1)$.
And you can check that you get analogue results
with the other three lines obtained by the combination of
$\pi _{1,2,x} $ and $ \pi _{2,3,y} $.
Best Answer
Simply write $$\left\{ \begin{array}{rcl} 4+t&=&6+2s\\ 5+t&=&11+4s\\ -1+2t&=&-3+s \end{array} \right.$$
And then solve the system formed by the two fisrt equations.
If the solution satisfies the third one, then the lines intersect and its intersection can be found substituting the found value for $t$ in the first line, or the value for $s$ in the second line.
If the solution doesn't satisfies the third equation, the lines do not intersect.