Answer
$\dfrac{x-1}{-1}=\dfrac{y-1}{1}=\dfrac{z-\dfrac{\pi}{4}}{1}$
Work Step by Step
Since, $\gamma =(\sqrt 2 \cos t,\sqrt 2 \sin t, t )$
The line which is tangent to the helix at the point $(1,1, \dfrac{\pi}{4})$ can be expressed as:
Now, $\dfrac{d \gamma}{dt} =\dfrac{d(\sqrt 2 \cos t,\sqrt 2 \sin t, t )}{dt}=(-\sqrt 2 \sin t,\sqrt 2 \cos t, 1 )$
$\implies \dfrac{d \gamma}{dt} (\dfrac{\pi}{4})=(-\sqrt 2 \sin (\dfrac{\pi}{4}),\sqrt 2 \cos (\dfrac{\pi}{4}), 1 )=(-1,1,1)$
Now, we need to write the parametric equations:
$\dfrac{x-1}{-1}=\dfrac{y-1}{1}=\dfrac{z-\dfrac{\pi}{4}}{1}$
