For what value of $K$ are the points $A(1, -1)$, $B(K/2, K/3)$, and $C(4, 1)$ collinear? 
For what value of $K$ are the points $A(1, -1)$, $B(K/2, K/3)$, and $C(4, 1)$ collinear?

The equation of line $AC$ is $2x-3y=5$. But the point $B$ doesn't satisfy the equation.
 A: 
"The equation of line AC is 2x-3y=5. But the point B doesnt satisfy the equation."

WHy not?
If that is the equation of the line and if we are told that $(\frac K2, \frac K3)$ is a point of the line, then it must be true that $2*\frac K2 - 3*\frac K3 = 5$.  For what value of $K$ is that true?
1) Is that the equation of the line?
The slope of the equation of the line is $m = \frac {C_y - A_y}{C_x - A_x} = \frac {1-(-1)}{4-1} = \frac 23$  so the equation is $(y-A_y) = m*(x-A_x)$ so $y-(-1) = \frac 23(x-1)$ or $y = \frac 23 x -\frac 53$ or as you put it $2x - 3y = 5$.
2) What value of $K$ satisfies $2*\frac K2 - 3*\frac K3 = 5$ ?
$K - K = 5$
$0 = 5$ 
There is no value of $K$ that satisfies.
So the answer to "For what value of K are the points A(1, -1), B(K/2, K/3), and C(4, 1) collinear?" is...
"None".
Which is a perfectly acceptable and valid answer.  Just because a textbook asks a question doesn't mean that there is an valid answer.  But in this case "none" is a valid answer.  Just because a textbook asks "when does this occur" doesn't mean the answer isn't "never".
A: If can you can proof that no real number $K$ will lead to a point $B$ which is on the line through $A$ and $C$, that is fine. 
It might be an error in the book or for educational purposes.
My attempt:
The line through $A$ and $C$ is
$$
(1-\lambda) A + \lambda C \quad (\lambda \in \mathbb{R})
$$
We equate with point $B$ and get
$$
(1 - \lambda) (1, -1) + \lambda (4,1) = (K/2, K/3)
$$
which gives the system
$$
(1 - \lambda) + 4 \lambda = K/2 \\
-(1-\lambda) + \lambda = K/3
$$
which is equivalent to the inhomogeneous system in unknowns $\lambda$ and $K$:
$$
\left[
\begin{array}{rr|r}
3 & -1/2 & -1 \\
2 & -1/3 & 1
\end{array}
\right]
\to
\left[
\begin{array}{rr|r}
1 & -1/6 & -2 \\
2 & -1/3 & 1
\end{array}
\right]
\to
\left[
\begin{array}{rr|r}
1 & -1/6 & -2 \\
0 & 0 & 5
\end{array}
\right]
$$
We subtract the second row from the first row, then subtract two times the first row from the second row. The resulting last row is inconsistent ($0\cdot \lambda + 0 \cdot K = 5$), so there is no solution.
A: The direction vector of the straight line $K(1/2,1/3)$ is obviously parallel to the direction vector $C-A=(3,2)$ of the straight line through $A$ and $C$.
