Answer
$x=1-t,y=t,z=1-t$
Work Step by Step
Given: $x=e^{-t} \cos t, y=e^{-t} \sin t,z=e^{-t} $
Equation can be written as: $r(t)= e^{-t}\cos t i+e^{-t}\sin t j+e^{-t} k $
Tangent vector $r'(t)$ can be calculated by taking the derivative of $r(t)$ with respect to $t$.
This yields,$r'(t)=e^{-t}(-\cos t- \sin t) i+e^{-t}(-\sin t+ \cos t)-e^{-t} k $
or, $r'(0)=e^{-0}(-\cos(0)- \sin(0)) i+e^{-0}(-\sin (0)+ \cos (0))-e^{-(0)} k =-i+j-k$
As we are given that the equation of tangent line passing through the points $(1,0,1)$, thus
$r(t)=\lt 1-t,t,1-t \gt$
Hence, the desired parametric equations of the tangent line are:
$x=1-t,y=t,z=1-t$