Finding Homology of Koszul Complex Let $R = \mathbb{Q}[x,y,z]$ and $I = \langle x,y,z\rangle$. I have the free resolution of $R/I$ as

$K : 0 \rightarrow R \rightarrow R^3 \rightarrow R^3 \rightarrow R \rightarrow 0$

I need to find the $H_1(K)$ which is equal to $\ker (\phi_1)/ \operatorname{im} (\phi_2)$.

But I am stuck here. I know the maps explicitly but I am not being able to find what is it exactly. I am guessing as this is a regular sequence the homology would be zero, so that sequence would be exact. But how to proceed ?
Theorem [Commutative Ring Theory-H Matsumura] (see Page-128)
  
  Let $A$ be a ring, $M$ an $A$-module, and $x_1,\ldots,x_n$ an
  $M$-sequence; then $H_p(x,M)=0$ for $p &gt; 0$ and $H_0(x,M)=M/xM$.
Theorem [Commutative Ring Theory-H Matsumura] (see Page-128)
  
  Let $A$ be a ring, $M$ an $A$-module, and $x_1,\ldots,x_n$ an
  $M$-sequence; then $H_p(x,M)=0$ for $p &gt; 0$ and $H_0(x,M)=M/xM$.