Answer
$x=2+ \dfrac{t}{2},y=ln 4+ \dfrac{t}{2},z=1+t$
Work Step by Step
From the given points we have $r(t)= \sqrt {t^2+3} i+ln(t^2+3) j+t k $
In order to find tangent vector $r'(t)$ we will have to take the derivative of $r(t)$ with respect to $t$.
Thus, $r'(t)=\dfrac{t}{ \sqrt {t^2+3}}i+\dfrac{2t}{ t^2+3}j+1 k $
or, $r'(t)=\lt \dfrac{t}{ \sqrt {t^2+3}},\dfrac{2t}{ t^2+3},1 \gt $
$r'(1)=\lt \dfrac{1}{ \sqrt {(1)^2+3}},\dfrac{2(1)}{ (1)^2+3},1 \gt=\lt \dfrac{1}{2}, \dfrac{1}{2},1 \gt$
Equation of tangent line passing through the points $(2,ln 4,1)$ is:
$r(t)=\lt (2+ \dfrac{t}{2}),(ln 4+ \dfrac{t}{2}), (1+t) \gt$
Therefore, the parametric equations of the line are:
$x=2+ \dfrac{t}{2},y=ln 4+ \dfrac{t}{2},z=1+t$
