Checking arithmetic for Intersection of two lines

Question

Let $L$ be the line with vector equation $$(x,y,z) = (1,3,7) + t(2,0,3),\ t\ \in \ \Bbb R$$ and let $M$ be the line with Cartesian equations $$\frac{x-3}{2}=y-5=z-6$$

Find the point of intersection of $L$ and $M$, or show that they do not intersect.

The answer given says there is no intersection but I have found one. I'm not sure who's wrong (probably me).

My working: rewriting $M$

$$(x,y,z)=(3,5,6)+s(2,1,1), \ s \ \in \ \Bbb R$$

If there is an intersection it can be found by making the $x$, $y$ and $z$ elements of equation $L$ and $M$ equal each other.

$$x: \ 3+2s=1+2t$$ $$y: \ 5+s=3 $$ $$z: \ 6+s=7+3t$$

Solving to give $s=-2$, $\ t=-1$ and the intersection point being $(-1,3,4)$. Is this correct?

Answer 1

There is no need to rewrite the second line in the same form:

We have one parameter $t$ to solve for in the first equation. So we know $x=1+2t, y=3, z=7+3t$.

Substitute those in the combined equations $\frac{x-3}{2}=y-5=z-6$ does give a solution: $z-6 = y-5$ becomes $1+3t= -2$, so $3t=-3$ hence $t=-1$, and $\frac{x-3}{2}=y-5$ becomes $\frac{2t-2}{2}=t-1 = -2$ so $t=-1$. So we get $(x,y,z)=(-1,3,4)$ and this lies on $L$ and on $M$ too as $-2 = -2 = -2$.

Answer 2

To check your answer,
for $t=-1$, $$(1,3,7)+(-1)(2,0,3)=(-1,3,4)$$ which is a point on $L$.
Also, you can check that $$\frac{(-1)-3}{2}=3-5=4-6$$ Hence $(-1,3,4)$ is also a point on $M$.
So you are correct, $(-1,3,4)$ is a point of intersection of line $L$ and $M$.