How to find a minimal polynomial (field theory)
To begin, clear denominators: $$(1 - \sqrt{5} + \sqrt{7}) \alpha = 3 \sqrt{5} - 2 \sqrt{7} + \sqrt{35}$$ We need to make the coefficient of $\alpha$ rational, so use a difference-of-squares trick to get rid of the $\sqrt{7}$ on the LHS (i.e. multiply both sides by $1 - \sqrt{5} - \sqrt{7}$), $$((1 - \sqrt{5})^2 - 7) \alpha = (3 \sqrt{5} - 2 \sqrt{7} + \sqrt{35})(1 - \sqrt{5} - \sqrt{7})$$ and after expanding and collecting like terms: $$(1 + 2 \sqrt{5}) \alpha = 1 + 4 \sqrt{5} + 7 \sqrt{7}$$ Now do the same again to deal with the $\sqrt{5}$ on the LHS: $$19 \alpha = 39 - 2 \sqrt{5} - 7 \sqrt{7} + 14 \sqrt{35}$$ Next, we have to deal with the irrational numbers on the RHS. First, we deal with $\sqrt{5}$ (and $\sqrt{35}$): move all the other terms over to the LHS, and square the resulting equation, $$(19 \alpha - 39 + 7 \sqrt{7})^2 = (-2 + 14 \sqrt{7})^2 \cdot 5$$ which expands to this: $$361 \alpha^2 - 1482 \alpha + 266 \sqrt{7} \alpha + 1864 - 546 \sqrt{7} = 6880 - 280 \sqrt{7}$$ To finish off, we deal with $\sqrt{7}$: put all multiples of $\sqrt{7}$ on the RHS and all others on the LHS, and then square the resulting equation: $$(361 \alpha^2 - 1482 \alpha - 5016)^2 = (- 266 \alpha + 266)^2 \cdot 7$$ Note that $19$ divides all the coefficients, so we can cancel that common factor: $$(19 \alpha^2 - 78 \alpha - 264)^2 = (-14 \alpha + 14)^2 \cdot 7$$ Finally, we obtain, $$361 \alpha^4 - 2964 \alpha^3 - 3984 \alpha^2 + 41184 \alpha + 69696 = 1372 \alpha^2 - 2744 \alpha + 1372$$ which simplifies to the desired equation: $$19 \alpha^4 - 156 \alpha^3 - 280 \alpha^2 + 2312 \alpha + 3596 = 0$$
$$\alpha-\alpha\sqrt{5}+\alpha\sqrt{7}=3\sqrt{5}-2\sqrt{7}+\sqrt{35},$$ $$\alpha-\sqrt{35}=(\alpha+3)\sqrt{5}-(\alpha+2)\sqrt{7},$$ $$(\alpha-\sqrt{35})^2=[(\alpha+3)\sqrt{5}-(\alpha+2)\sqrt{7}]^2,$$ $$\alpha^2+35-2\alpha\sqrt{35}=5(\alpha+3)^2+7(\alpha+2)^2-2\sqrt{35}(\alpha+2)(\alpha+3),$$ $$2\sqrt{35}(\alpha^2+4\alpha+6)=11\alpha^2+58\alpha+38,$$ $$[2\sqrt{35}(\alpha^2+4\alpha+6)]^2=(11\alpha^2+58\alpha+38)^2.$$ From the last equality, we get that minimal polynomial is $19x^4-156x^3-280x^2+2312x+3596$.
A general purpose method is that the equation
$$ \sum_{k=0}^n c_k \alpha^k = 0 $$
is a linear equation in the unknowns $c_k$, and thus this can be solved with linear algebra.
Since the number itself is a rational linear combination of the four linearly independent numbers $1, \sqrt{5}, \sqrt{7}, \sqrt{35}$, we get "$4$ equations in $n+1$ unknowns", so $n=4$ will guarantee a solution exists.
If desired, you can avoid computing the quotient by using the fact $$ \sum_{k=0}^n c_k \left(\frac{\mu}{\nu}\right)^k = 0 \quad \Longleftrightarrow\quad \sum_{k=0}^n c_k \mu^k \nu^{n-k} = 0 $$