Computing a regular sequence of generators for an ideal

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Let $R = \mathbb{C}[x_1,\ldots,x_n]$. Let $I$ be an ideal, and suppose we know a finite list of generators for $I$, say $I = \langle f_1,\ldots,f_k\rangle$.

Is this information enough to compute a new generating set $g_1,\ldots,g_\ell$ for $I$ such that $(g_1,\ldots,g_\ell)$ is a (strongly) regular sequence? If not, then what about if the $f_i$ are assumed to be irreducible and homogeneous?

Context: algebraic geometry, algebraic K-theory. I'm trying to use Koszul complexes to compute Grothendieck classes of some varieties, but I'm having trouble coming up with regular sequences (which I need for a Koszul resolution).

Specific case: $R = \mathbb{C}[x,y,z,r,s,t]$, $I = \langle f,g,h\rangle$ where $$ f = xr - yz, ~ g = xt - ys, ~ h = zt - rs. $$ These are the $2\times 2$ minors of the matrix $$ \begin{bmatrix} x & y\\ z & r\\ s & t \end{bmatrix}. $$ I'd like a new set of generators for $I$ that is a regular sequence, if one exists. The current set doesn't work because any two polynomials are a regular sequence, but the third is then a zero divisor in the quotient: e.g. $zg - sf = xh$.

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First, let me address the general question of ideals being generated by regular sequences: as Youngsu pointed out, such ideals are called complete intersections. In a Cohen-Macaulay ring (which the polynomial ring is), these are the same as ideals of the principal class - i.e. ideals which can be generated by $\DeclareMathOperator{\ht}{ht}$$\ht(I)$-many elements.

Now in your specific case, $I$ is the ideal of $2 \times 2$ minors of a generic $3 \times 2$ matrix. Such an ideal has height $2$ (e.g. by the Eagon-Northcott bound). As it happens, $I$ is Cohen-Macaulay (meaning $R/I$ is CM) but $I$ is not a complete intersection. One way to see this is via not the Koszul complex, but another famous complex, the Hilbert-Burch resolution:

$$0 \to R^2 \xrightarrow{\varphi} R^3 \xrightarrow{\psi} R \to R/I \to 0$$

where $\psi$ is the matrix $\begin{bmatrix} f & g & h \end{bmatrix}$ and $\varphi = \begin{bmatrix} s&t\\ -z&-r\\ x&y \end{bmatrix}$ is the matrix of relations among $f, g, h$ resulting from the equations $sf - zg + xh = 0 = tf - rg + yh$ - notice that the failure of $f, g, h$ to form a regular sequence gave rise exactly to the matrix $\varphi$. One immediately sees that this resolution is minimal (since all the entries of each map lies in the homogeneous maximal ideal), which implies that $I$ requires at least $3$ generators, and thus is not a complete intersection.