Is this variant of ATM decidable?
You can show that $\mathrm{BTM}$ is undecidable with a simple reduction from $\mathrm{ATM}$.

Let $\langle M,w\rangle \in ATM$. We define the reduction function $f$,  $\mathrm{HP} \leq \mathrm{BTM}$ as follow:

$f(\langle M,w\rangle) = M_w $, Where $M_w$ on every input $x$, runs $M$ with input $w$. if $M$ stopped, then $M_w$ accepts $x$. if $M$ rejects, then $M_w$ rejects. Clearly, if $\langle M,w\rangle \in ATM$ then $L(M_w) = \Sigma^*$, and in particular $101 \in L(M_w)$. if $\langle M,w\rangle \notin ATM$ then   $L(M_w) = \emptyset$, and $101 \notin L(M_w)$.

Because $\mathrm{ATM} $is undecidable, so is $\mathrm{BTM}$.

It should be intuitive that $\mathrm{BTM}$ is undeciadble, because given a turing machine $M$, you can't tell if $M$ will halt with input $101$