Answer
The minimum area is 1.88.
Work Step by Step
Let $P(x, y, 0)$ be a point on the curve $\ln x=y$ in the yx plane. Let $u$ be the vector from $P$ to the point (2,-1,0) , and let $v$ be the vector from $P$ to the point $(3,2,2) .$ And then:
\[
(-x+2) \vec{i}+(-\ln x-1) \vec{j}=\vec{u}
\]
\[
(-x+3) \vec{i}+(-\ln x+2) \vec{j}+2 \vec{k}=\vec{v}
\]
The area of the triangle is given by
\[
\|\vec{u} \times \vec{v}\|(1 / 2)=A
\]
Calculation of the cross product gives:
\[
\vec{u} \times \vec{v}=(-2-2 \log (x)) i+(-4+2 x) j+(7+\log (x)-3 x) k
\]
So:
\[
\frac{1}{2} \sqrt{(1+\log (x))^{2}4+(7-3 x+\log (x))^{2}+(-2+x)^{2}4}=A
\]
We solve for $x$ with Mathematica to find the minimum area:
\[
d A / d x=\frac{11+(-32+13 x)x+(-3 x+5) \log (x)}{2 x \sqrt{x(13 x-58)+69+\log (x)(-6 x+5 \log (x)+22)}}=0
\]
We obtain $x=2.09158 .$ So the minimum area is 1.88.
