Answer
$\langle \frac{-6}{\sqrt{6}},\frac{12}{\sqrt{6}}, \frac{6}{\sqrt{6}}\rangle$
Work Step by Step
Let $\mathbf v=\langle -2,4,2 \rangle$. We can obtain a unit vector in the same direction as $\mathbf v$ by multiplying $\mathbf v$ by $\frac{1}{|\mathbf v|}$. Then we can multiply that unit vector by $6$, and this will give us a vector with length $6$ in the same direction as $\mathbf v$.
So first, we find the unit vector in the same direction as $\mathbf v$.
By definition, $|\mathbf v|=\sqrt{(-2)^2+4^2+2^2}=\sqrt{4+16+4}=\sqrt{24}=2\sqrt{6}$.
So $\frac{\mathbf v}{|\mathbf v|}=\frac{\mathbf v}{2\sqrt{6}}=\frac{\langle -2,4,2 \rangle}{2\sqrt{6}}=\langle \frac{-2}{2\sqrt{6}},\frac{4}{2\sqrt{6}}, \frac{2}{2\sqrt{6}}\rangle =\langle \frac{-1}{\sqrt{6}},\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}\rangle$ is a unit vector in the same direction as $\mathbf v$.
Now we multiply this unit vector by 6 to obtain our desired vector:
$6\langle \frac{-1}{\sqrt{6}},\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}\rangle=\langle \frac{-6}{\sqrt{6}},\frac{12}{\sqrt{6}}, \frac{6}{\sqrt{6}}\rangle$.
