The straight line kx-y + 1-3k = 0, when k changes, what is the point where all the straight lines are constant

The straight line kx-y + 1-3k = 0, when k changes, what is the point where all the straight lines are constant

K（X-3）=Y-1
The fixed point is (3,1)

The straight line kx-y + 1 = 3k, when k changes, all the straight lines pass through the fixed point ()
1.k(x-3)=y-1
2. If x-3 = Y-1 = 0, the equation holds regardless of the value of K
3. So x = 3, y = 1
4. All the straight lines pass through the fixed point (3,1). How did the first step come about? The second step is not clear,

kx-y+1=3k
kx-y+1-3k=0
k(x-3)=y-1
Because no matter what value K takes, it passes through the fixed point
x-3=0
Then Y-1 = 0
Namely
Over (3,1)

When k changes, all the lines pass through the vertex____ ?

The intercept formula of the linear equation is y = K (x-3) + 1
So when k = 3
K can be any value
It is found that the line crosses the point (3,1)

The intersection of the line y = - x + 3K and the line y = x-7k is on the line 2x + 3Y = 6

y=-x+3k
y=x-7k
So - x + 3K = x-7k
x=5k
y=x-7k=-2k
On 2x + 3Y = 6
10k-6k=6
k=3/2