Answer
$\dfrac{\sqrt{30}}{18}$
Work Step by Step
Given: $r(t)=\lt t^2, \ln t, t \ln t \gt$
To calculate the curvature of the curve we will have to use Theorem 10 such as: $\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}$
Thus,
$r'(t)=\lt 2t, \frac{1}{t}, \ln t +1 \gt$
and $r''(t)=\lt 2, -\frac{1}{t^2}, \frac{1}{t} \gt$
Now,
$\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}=\dfrac{|\frac{1}{t^2}(2+\ln t)i+2 \ln t j4/tk^ |}{[4t^2+\frac{1}{t^2}+ (\ln t+1)^2]^{3/2}}$
At point $(1,0,0)$
$\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}=\dfrac{|2 i+0 j-4k |}{[6]^{3/2}}=\dfrac{\sqrt{30}}{18}$
Hence, $\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}|_{(1,0,0)}=\dfrac{\sqrt{30}}{18}$
