To find the constraint equations use the fact that constraints do no work, and thus the product of constraint forces $\vec{F}$ and velocity $\vec{v}$ must be zero in a vectorial form $$ \vec{F} \cdot \vec{v} = 0 $$
Where $\cdot$ is the vector inner product.
We only care about directions here, and the direction of velocity is found by differentiation $$ \vec{v} = \frac{\rm d}{{\rm d}t} \vec{r}(t) = \pmatrix{1 \\ t \cos(t) + \sin(t) \\ 2 t \exp(t^2)} \dot{t} $$
Where $\vec{r}(t) = \pmatrix{t \\ t \sin(t) \\ \exp(t^2) }$ is the path curve equation.
So the constraint equations are all the directions orthogonal to $\vec{v}$. For example
$$\vec{n}_1 = \vec{v} \times \pmatrix{0\\0\\1} \propto \pmatrix{t \cos(t)+\sin(t) \\ -1 \\ 0}$$
$$\vec{n}_2 = \vec{v} \times \pmatrix{0\\1\\0} \propto \pmatrix{-2t \exp(t^2)\\0\\1}$$
Where $\times$ is the vector cross product.
So the constraint forces are some kind of linear combination of the two constraint directions
$$ \vec{F} = \lambda_1 \pmatrix{t \cos(t)+\sin(t) \\ -1 \\ 0} + \lambda_2 \pmatrix{-2t \exp(t^2)\\0\\1} $$
You can prove that $\vec{F} \cdot \vec{v}$ is zero for all instances of $t$. So for ever $t$ there are two variables that describe the constraints, $\lambda_1$ and $\lambda_2$.