Show that x = 2 is one of the root of below matrix

{ x -6 1 }
{ 2 -3x x-3 }
{ -3 2x x+2} = 0

I'm going to assume that you are telling us that the determinant is 0.

[x (-3x)(x + 2)] + [2 * 2x * 1] + [6(x - 3) * 3] - [1 * 3x * 3] + [ 2 * 6 (x + 2)] - x * 2x(x - 3)] = 0

-3x²(x + 2) + 4x + 18(x - 3) - 9x + 12(x + 2) - 2x²(x - 3) = 0

-3x³ - 6x² + 4x + 18x - 54 - 9x + 12x + 24 - 2x³ + 6x² = 0

-5x³ + 25x - 30 = 0
x³ - 5x + 6 = 0

Check the information you gave. This does not have a root = 2.