Find the value of k for which the points (3,2), (1,2), (5,k) are collinear

The correct option is 2 (which corresponds to the value $k=2$).

To understand why this is correct, let us analyze the condition for three points to be collinear.
Three points, say $A_{ }(x_1,y_1)$, $B_{ }(x_2,y_2)$, and $C_{ }(x_3,y_3)$, are collinear if they all lie on the same straight line. This means that the area of the triangle formed by these three points must be equal to zero.

The formula for the area of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$ is given by:
$$\text{Area}=\frac{1}{2}\left|x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right|$$

For collinear points, we set this area to 0:
$$x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0$$

Here, the given points are:
$(x_1,y_1)=(3,2)$
$(x_2,y_2)=(1,2)$
$(x_3,y_3)=(5,k)$

Substituting these values into the collinearity condition:
$$3(2-k)+1(k-2)+5(2-2)=0$$

Now, let us simplify each term step-by-step:
First term: $3(2-k)=6-3k$
Second term: $1(k-2)=k-2$
Third term: $5(2-2)=5\left(0\right)=0$

Adding these simplified terms together:
$$(6-3k)+(k-2)+0=0$$

Combine the constant terms and the terms involving $k$:
$$4-2k=0$$

Rearranging the equation to solve for $k$:
$$2k=4$$
$$k=\frac{4}{2}$$
$$k=2$$

Alternatively, we can also observe the y-coordinates of the first two points: both $(3,2)$ and $(1,2)$ lie on the horizontal line $y=2$. For a third point $(5,k)$ to be on the same line, its y-coordinate must also be $2$, which immediately gives $k=2$.

Therefore, the value of $k$ for which the points are collinear is 2.