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is there anyone who knows how to (can?) solve it. Any hints or directions would be helpful.

Write the formula for the distance from a point to a line L. Then calculate the distance of the point $P_1(1,2,4)$ to the line L which is the intersection of two planes $x-y + 2z = 1$ and $x + 3y-z = 4$ .

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    "Then calculate the distance of the point $P_1(1,2,4)$ and the distance along which intersect the $x-y + 2z = 1$ and $x + 3y-z = 4$ planes." I don't understand this sentence. Are you sure you've copied it down correctly? – 2017-02-08
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    @Théophile It's about calculating the distance from line L where along the side of this line two planes are crossing. – 2017-02-08
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    A better way of saying that, is, if I understand you correctly: The line $L$ is the intersection of two planes. – 2017-02-08
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    @BadamBaplan Yes, that is what I meant :) – 2017-02-08

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Say you have a point $P$ and some line $L$ with direction vector $\vec v$, and you know $P_2$ is on the line. We are interested in the distance from $P$ to $L$. Then we construct a vector going from $P_2$ to $P$, that is $\vec P-\vec P_2$.

The magnitude of cross product $|(\vec P-\vec P_2) \times \vec v|$ gives the area of the parallelogram formed by these two vectors. The distance we are seeking is the height of this parallelogram. The base is of length $|\vec v|$. And $\text{Area}=(\text{base}) (\text{height})$. So the distance is,

$$\text{height}=\frac{\text{Area}}{\text{base}}=\frac{|(\vec P-\vec P_2) \times \vec v|}{|\vec v|}$$

enter image description here

First let $z=0$ then from your two planes we have,

$$x-y=1$$

$$x+3y=4$$

So then $x=1.75$ and $y=0.75$. So one choice of $\vec P_2$ is $\langle 1.75,0.75,0 \rangle$. A direction vector of the line from the intersection of your two planes is the cross product of the normals of the two planes.

$$\vec v=\langle 1,-1,2 \rangle \times \langle 1,3,-1 \rangle=\langle -5,3,4 \rangle$$

So then the distance is,

$$d=\frac{|(\langle 1,2,4 \rangle-\langle 1.75,0.75,0 \rangle) \times \langle -5,3,4 \rangle|}{| \langle -5,3,4 \rangle|}$$

$$d=\frac{|\langle -0.75,1.25,4 \rangle \times \langle -5,3,4 \rangle|}{\sqrt{50}}$$

$$d=\frac{|\langle -7,-17,4 \rangle|}{\sqrt{50}}$$

$$=\sqrt{\frac{354}{50}}$$

$$=\frac{\sqrt{177}}{5}$$

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    do you know how should I use it? – 2017-02-11
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    I will post the full answer @DaveM – 2017-02-11
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    @Ahmend S. Atalla thanks, why you are making z=0 ? – 2017-02-11
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    I just would like to find one point on the line. z=0 will do, or anything like that. – 2017-02-12
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Hint:

The projection of $P$ to $L$ (let $Q$) is the intersection of a plane perpendicular to $L$, through $P$. This plane is perpendicular to any plane that contains $L$ and in particular the two given planes.

Then the direction of the normal to the plane is the cross product of the normals, $$(1,-1,2)\times(1,3,-1)=(-5,3,4)$$

so that the equation of the plane is

$$-5(x-1)+3(y-2)+4(z-4)=0.$$

Now you solve the $3\times3$ system formed by the three plane equations,

$$\begin{cases}-5x+3y+4z=17,\\x-y + 2z = 1,\\x + 3y-z = 4,\end{cases}$$

giving $Q$, and you return the distance $PQ$.

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    And result of the equation is 148/25 ? – 2017-02-09
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    @DaveM: no idea, I didn't compute. If you post your $x,y,z$, I can check them. – 2017-02-09
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    $$ x = -\frac{3}{5}$$ $$ y = \frac{54}{25}$$ $$ z = \frac{47}{25}$$ I solved the system of equations by the method of Gauss-Jordan . – 2017-02-09
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    @DaveM: from the plane equation. – 2017-02-09
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    Ahh... sorry for that I'm just tired :) . So are x,y,z correct or i did some stupid mistake again xD ? – 2017-02-09
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    @DaveM: no, they don't fit in the equations. – 2017-02-09
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    But wolfram thinks opposite when I put x,y,z into three equations they are giving me 17,1,4 because wolfram returns true – 2017-02-09
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    @DaveM: sorry, I made a sign error. But why do you ask me a check if you can do it with WA ??? – 2017-02-09
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    Cause I wanted to be sure that I'm doing it the way that you said me :) . And I always compute myself, wolfram is for the doubts. So if x,y,z are true the distance 148/25 should be true ? – 2017-02-09
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The distance from a point (m, n) to the line Ax + By + C = 0 is given by:

perpendicular distance formula

$$d=$$ $$\frac{|Am+Bn+C|}{\sqrt{A^2+B^2}}$$

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    I'm trying and trying and i still have no idea how to use it :( – 2017-02-10
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    But the line is in 3D not the xy plane – 2017-02-11