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Let $\mathbf u = [3,-1,2,5,0]$, and v = $[1,0,-2,1,4]$.

3u = $[9, -3, 6, 15, 0]$

2v = $[2, 0, -4, 2, 8]$

3u - 2v = $[9, -3, 6, 15, 0]$ - $[2, 0, -4, 2, 8]$ = $[7, -3, 10, -13, 8]$

$$\|[7, -3, 10, -13, 8]\| = \sqrt{(7)^2 + (-3)^2 + (10)^2 + (-13)^2 + (8)^2} = \sqrt{391}$$

This is one answer, and I think it is correct, but there is also a law that says:

$\|xu\| = |x|\cdot\|u\|$, where $x$ is a constant, $u$ is a vector ($\|3u\| = |3|\cdot\|u\|$)

So,

$$\|3u - 2v\| = 3\|u\| - 2\|v\| = 3\sqrt{39} - 2\sqrt{22}$$

These two answers do not agree. I think the second one is false, but then how is the law wrong in this case?

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    In general, $\|x+y\| \le \|x\| + \|y\|$, where $\|\cdot\|$ is a norm. This property is called the triangle inequality. – 2017-01-14

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You're incorrect in the second portion, with your mistake being that you are assuming that $$||3u - 2v|| = ||3u|| + ||-2v||$$ This is not true, as it is not true that $||x+y||=||x||+||y||$. Instead, it should be said that $$||3u - 2v|| \le ||3u|| + ||-2v||$$ as shown here.