If the function $$f(x) = \begin{cases} a|\pi - x| + 1, & x \le 5 \\ b|x - \pi| + 3, & x \gt 5 \end{cases}$$ is continuous at $$x = 5$$, then the value of $$a - b$$ is:

We have the piece-wise definition

$$f(x)=\begin{cases} a\,|\pi-x|+1, & x\le 5\\[4pt] b\,|x-\pi|+3, & x\gt 5 \end{cases}$$

For continuity at a point, the following condition must hold:

$$\lim_{x\to 5^-}f(x)=f(5)=\lim_{x\to 5^+}f(x).$$

First we calculate the actual value of the function at $$x=5$$. Since $$5\le5$$, the first branch is used:

$$f(5)=a\,|\pi-5|+1.$$

Because $$\pi\approx3.14\lt 5$$, the quantity $$\pi-5$$ is negative, so

$$|\pi-5|=5-\pi.$$

Substituting this into the expression for $$f(5)$$, we obtain

$$f(5)=a\,(5-\pi)+1.$$

Next we find the left-hand limit as $$x\to5^-$$. For $$x\lt 5$$ we again use the first branch, and the argument of the absolute value, $$\pi-x$$, is still negative (because $$x\lt 5\lt \pi?$$ No, $$\pi\lt 5$$ so $$\pi-x\lt 0$$), giving

$$\lim_{x\to5^-}f(x)=\lim_{x\to5^-}\bigl(a\,|\pi-x|+1\bigr)=\lim_{x\to5^-}\bigl(a\,(5-\pi)+1\bigr)=a\,(5-\pi)+1.$$

Thus the left-hand limit already equals $$f(5)$$, as expected.

Now we evaluate the right-hand limit. For $$x\gt 5$$ we use the second branch: $$f(x)=b\,|x-\pi|+3.$$ When $$x\gt 5\gt \pi$$, the difference $$x-\pi$$ is positive, hence

$$|x-\pi|=x-\pi.$$

Taking the limit as $$x\to5^+$$ gives

$$\lim_{x\to5^+}f(x)=\lim_{x\to5^+}\bigl(b\,(x-\pi)+3\bigr)=b\,(5-\pi)+3.$$

Continuity demands equality of the two one-sided limits, so we equate:

$$a\,(5-\pi)+1=b\,(5-\pi)+3.$$

Bringing all terms involving $$a$$ and $$b$$ to one side and constants to the other side, we write

$$a\,(5-\pi)-b\,(5-\pi)=3-1.$$

Factoring out the common factor $$5-\pi$$ on the left gives

$$(5-\pi)\,(a-b)=2.$$

Finally, solving for the desired difference $$a-b$$, we divide both sides by $$5-\pi$$:

$$a-b=\frac{2}{5-\pi}.$$

This matches Option A.

Hence, the correct answer is Option A.