Show that the vectors (3, 0, -3), (-1, 1, 2), (4, 2, -4), and (2, 1, 1) are linearly dependent over ℝ.

What you need to show

To prove these vectors are linearly dependent, you just need to find one set of real numbers, not all zero, such that $$a(3,0,-3)+b(-1,1,2)+c(4,2,-4)+d(2,1,1)=(0,0,0).$$

Turn the vector equation into a system

Match coordinates component-by-component: $$3a-b+4c+2d=0$$ $$b+2c+d=0$$ $$-3a+2b-4c+d=0$$

Solve the homogeneous system

From the second equation, $$b=-2c-d.$$ Substitute into the first: $$3a-(-2c-d)+4c+2d=0$$ $$3a+6c+3d=0 \,\Rightarrow\, a+2c+d=0 \,\Rightarrow\, a=-2c-d.$$ Now substitute $a=-2c-d$ and $b=-2c-d$ into the third equation: $$-3(-2c-d)+2(-2c-d)-4c+d=0$$ $$6c+3d-4c-2d-4c+d=0$$ $$-2c+2d=0 \,\Rightarrow\, d=c.$$ Then $$a=-2c-d=-3c, \quad b=-2c-d=-3c.$$ Choose $c=1$ (any nonzero value works). Then $$a=-3,\; b=-3,\; c=1,\; d=1.$$

Write the dependence relation clearly

Plugging these coefficients back gives $$-3(3,0,-3) -3(-1,1,2) + (4,2,-4) + (2,1,1) = (0,0,0).$$ Since the coefficients are not all zero, the four vectors are linearly dependent over $\mathbb{R}$.