1. We need the limit

to exist, where $[t]$ denotes the greatest integer function.

2. First, simplify the numerator near $x=7$.

Let $x\to 7$. Then $1-x\to -6$.

We check from both sides:

• If $x\to 7^-$, write $x=7-h$ with $h>0$ small. Then $1-x=1-(7-h)=-6+h,$ where $-6+h\in (-6,-5)$ for small $h$. Hence $[1-x]=[-6+h]=-6.$

• If $x\to 7^+$, write $x=7+h$ with $h>0$ small. Then $1-x=1-(7+h)=-6-h,$ where $-6-h\in (-7,-6)$ for small $h$. Hence $[1-x]=[-6-h]=-7.$

So the numerator becomes:

• as $x\to 7^-$: $18-[1-x]=18-(-6)=24,$
• as $x\to 7^+$: $18-[1-x]=18-(-7)=25.$

3. Now examine the denominator: $[x-3a].$ As $x\to 7$, we have $x-3a\to 7-3a$.

For the limit to exist, since the numerator has different left and right values ($24$ and $25$), the denominator must also take different left and right values in such a way that the two quotient limits are equal.

This can happen only if $7-3a$ is an integer, because the greatest integer function jumps only at integers.

Since $a$ is an integer, $3a$ is an integer, so $7-3a$ is indeed an integer for every integer $a$.

Let $n=7-3a,$ where $n$ is an integer.

Then near $x=7$:

• if $x\to 7^-$, then $x-3a\to n^-$, so $[x-3a]=n-1=6-3a,$
• if $x\to 7^+$, then $x-3a\to n^+$, so $[x-3a]=n=7-3a.$

Therefore,

• Left-hand limit: $L^- = \frac{24}{6-3a},$
• Right-hand limit: $L^+ = \frac{25}{7-3a}.$

4. For the limit to exist, we need $\frac{24}{6-3a}=\frac{25}{7-3a}.$ Cross-multiplying: $24(7-3a)=25(6-3a).$ So, $168-72a=150-75a,$ $168-150=-75a+72a,$ $18=-3a,$ $a=-6.$

5. Verify: If $a=-6$, then $[x-3a]=[x+18].$ As $x\to 7$, $x+18\to 25.$ Then

• from left: $[x+18]=24$,
• from right: $[x+18]=25$.

Thus the quotient becomes

• left: $24/24=1$,
• right: $25/25=1$.

Hence the limit exists.

6. Therefore, $\boxed{a=-6}$ which corresponds to Option A.