The area of a triangle with vertices (-3, 0) (3, 0) and (0, k) is 9 sq. units. The value of k will be

The correct option is 3 (corresponding to the option "3" in the list).

To find the value of $k$, we can use the formula for the area of a triangle with given vertices. Let the vertices of the triangle be:
$(x_1,y_1)=(-3,0)$
$(x_2,y_2)=(3,0)$
$(x_3,y_3)=(0,k)$

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

Given that the area of the triangle is 9 square units, we substitute the coordinates of the vertices into the formula:
$9=\frac{1}{2}|-3(0-k)+3(k-0)+0(0-0\left)\right|$

Simplify the expression inside the absolute value:

Multiply both sides of the equation by 2:

Divide by 6:
$\left|k\right|=3$

Taking the positive value of the absolute expression matching the correct answer option, we have:
$k=3$