67 If $x>0, y>0, z>0$, and $x^{2}+y^{2}+z^{2}=1$, then the minimum value of $\frac{y z}{x}+\frac{x z}{y}+\frac{x y}{z}$ is
$67 \sqrt{3}$. Let $\frac{y z}{x}=a, \frac{x z}{y}=b, \frac{x y}{z}=c$, then
$$
x^{2}+y^{2}+z^{2}=1,
$$

becomes
$$
a b+b c+c a=1,
$$

so
$$
a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a=1 .
$$

thus we have
$$
(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+a c) \geqslant 3,
$$

which means
$$
\frac{y z}{x}+\frac{x z}{y}+\frac{x y}{z} \geqslant \sqrt{3},
$$
equality holds when $x=y=z$.