Let $\Lambda_A$ be a lattice with a basis $A= (a_1,a_2,a_3),$ where $a_1=(1,1,1)^T,a_2=(1,0,2)^T$, and $a_3=(0,1,1)^T$. Let $\Lambda_B$ be a lattice with a basis $B=(b_1,b_2,b_3)$, where $b_1=(1,2,6)^T,b_2=(3,1,3)^T,$ and $b_3=(3,3,7)^T$ Show that $\Lambda_B$ is a sublattice of $\Lambda_A$. Find the index of $\Lambda_B$ in$\Lambda_A.$
This is the way I approached this:
Let $\Lambda_o$ be a sublattice of a lattice $\Lambda$, then$[\Lambda:\Lambda_o]=|det\Lambda_o|/|det\Lambda|$.
$$ A=\begin{bmatrix} 1 & 1 & 0\\ 1 & 0 &1 \\ 1 & 2 & 1 \end{bmatrix} = \; 1\begin{bmatrix} 0 & 1\\ 2 & 1 \end{bmatrix} \; -1\begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix} + \; 0\begin{bmatrix} 0 & 1\\ 2 & 1 \end{bmatrix}$$ = det $A=-2$.
$$ B=\begin{bmatrix} 1 & 3 & 3\\ 2 & 1 &3 \\ 6 & 3 & 7 \end{bmatrix} = \; 1\begin{bmatrix} 1 & 3\\ 3 & 7 \end{bmatrix} \; -3\begin{bmatrix} 2 & 3\\ 6 & 7 \end{bmatrix} + \; 3\begin{bmatrix} 2 & 1\\ 2 & 3 \end{bmatrix}$$ = det $B=10$.
This would be a sublattice since the determinants are divisible. Did I take the right direction?
You have observed that $\det \Lambda_A$ divides $\det \Lambda_B$. Thus, if $\Lambda_A$ is a sublattice of $\Lambda_B$, then the index is $$[\Lambda_B : \Lambda_A] = |\det \Lambda_B| / |\det \Lambda_A| =5.$$ However, divisibility doesn't imply that $\Lambda_A \subset \Lambda_B$, i.e. you need to show that any $b_1,b_2,b_3$ can be expressed as a linear combination of $a_1,a_2,a_3$ with integral coefficients.
For example, the lattice $\Lambda$ spanned by $(1,1/2)$ and $(0,1)$ has determinant $1$. However, $\Lambda \not\subset \mathbb{Z}^2$, although it divides $\det \mathbb{Z}^2 =1$.
Edit: Let us show that the index is, in fact, the quotient of the determinants. (This argument can be also found in the below mentioned book of Cassels.)
Given a lattice $\Lambda_0$ and a sublattice $\Lambda_1 \subset \Lambda_0$, we can view both as groups. So $[\Lambda_1 : \Lambda_0]$ is the number of cosets. One standard fact is that there exists a basis $b_1, \ldots b_n$ of $\Lambda_1$ such that $a_i = \sum_{j=1}^i u_{i,j} b_j$ with some $u_{i,j} \in \mathbb{Z}$ is a basis. (See e.g. in the book 'Introduction to Geometry of Numbers' of Cassels.) Thus $$\det \Lambda_1 = \prod_{i=1}^n u_{i,i} \det \Lambda_0.$$ On the other hand any $$c= \sum_{i=1}^n n_i b_i$$ with $n_i \in \mathbb{N}$ and $n_i < |u_{i,i}|$ gives some different point in $\Lambda_0 \backslash \Lambda_1$ and this are all points, as can be checked. Thus, we find that $$[\Lambda_0 : \lambda_1| = \prod_{i=1}^n |u_{i,i}| = |\det \Lambda_0| / |\det{\Lambda_1}|.$$