Why can the determinant be assumed to be 0?

The text is not claiming that the determinant is $0$. The text is saying "Let's find out for which values of lambda the determinant is $0$!"

So the determinant is $\lambda^2 - 10\lambda + 30$, and you want to find the $\lambda$ such that it is equal to zero. What do you do? You set it equal to zero and solve for $\lambda$. That is, you solve the equation

$$\lambda^2 - 10\lambda + 30 = 0$$

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As for why you are interested in the values of $\lambda$ that make the determinant equal to $0$, remember that

$$rank(A-\lambda I) = n \iff det(A - \lambda I) \neq 0$$

So, if $det(A-\lambda I) \neq 0$, you will find that the only solution to $(A - \lambda I)x = 0$ is $x = 0$ (due to the fact that the rank of the matrix is full, hence the kernel only contains the $0$ vector). This means that the only $x$ such that $Ax = \lambda x$ is $x=0$, which means that $x$ is not an eigenvector.

So the only way to have eigenvectors is to have the determinant of $A - \lambda I$ be equal to zero, so that's why to find eigenvalues you look for the values of $\lambda$ that make $det(A - \lambda I) = 0$

For a square matrix like $M = (A - \lambda I)$, the equation $Mx = 0$ will have a non-zero solution $x$ if and only if $M$ doesn't have an inverse, which is true if and only if the determinant of $M$ is $0$.

The determinant of a $n\times n$ matrix $M$ is equal to $0$ if and only if the rank of the matrix is smaller than $n$, which happens if and only if the kernel of the matrix is non-empty, which happens if and only if there exists some vector $x\ne0$ such that $Mx=0$.

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Therefore, $\lambda$ is an eigenvalue of $A$ $\iff$ the determinant of $A-\lambda I$ is equal to $0$.