You need to solve the equation `A*X = 0` and then you need to decompose the result to find the requested basis, such that:

`A*X = 0`

`((9,-6,3),(-9,6,-3),(12,-8,4))*((x_1),(x_2),(x_3))` = ((0),(0),(0))

`((9x_1 - 6x_2 + 3x_3),(-9x_1 + 6x_2 - 3x_3),(12x_2 - 8x_2 + 4x_3)) = ((0),(0),(0))`

`{(9x_1 - 6x_2 + 3x_3 = 0),(-9x_1 + 6x_2 - 3x_3 = 0),(12x_2 - 8x_2 + 4x_3 = 0):}`

`Delta = 216 + 216 + 216 - 216 -216 - 216 = 0`

Since `Delta = delta = 0,` hence, you may consider `x_2,x_3` as free variables, such that:

`9x_1 = 6x_2 - 3x_3 => 3x_1 = 2x_2 - x_3`

Considering `x_2 = alpha, x_3 = beta` yields:

`3x_1 = 2alpha - beta => x_1 = (2alpha - beta)/3`

Hence, the solutions to equation `A*X = 0` is the following:

`X = (((2alpha - beta)/3),alpha, beta)`

Decomposing the solution yields:

`X = alpha*((2/3),(1),(0)) + beta*((-1/3),(0),(1))`

**Hence, the vectors `bar u = ((2/3),(1),(0))` and `bar v = ((-1/3),(0),(1))` represent a basis of the null space of A.**