It is obvious that a triangle with side lengths of $6, 8, 10$ is a right-angled triangle, with a perimeter of 24 and an area of 24 as well.

To prove that there is only one line that simultaneously bisects the perimeter and area of this triangle, we need to consider different scenarios of how this line intersects the sides of the triangle. Thus, the problem can be broken down into 4 sub-problems.
(1) When the line passes through a vertex of the triangle, can this line bisect both the perimeter and area of the triangle?
(2) When the line intersects the shorter leg and the hypotenuse, can this line bisect both the perimeter and area of the triangle?
(3) When the line intersects the longer leg and the hypotenuse, can this line bisect both the perimeter and area of the triangle?
(4) When the line intersects the two legs, can this line bisect both the perimeter and area of the triangle?
We will address these sub-problems one by one.
Let this triangle be $\triangle ABC$, with $\angle C=90^{\circ}$, $AC=6$, $BC=8$, and $AB=10$, and the line be $l$.
(1) When the line $l$ passes through a vertex of $\triangle ABC$, since any line passing through a vertex of a triangle and bisecting the area of the triangle must bisect the opposite side, it naturally cannot bisect the perimeter of the triangle. Therefore, this line does not meet the requirements of the problem.
(2) As shown in Figure 1, when the line $l$ intersects $AB$ and $AC$ at points $S$ and $T$ respectively, let $CT=x$ $(x>0)$, then $AT=6-x$. Clearly, the semi-perimeter is 12.

If the line $l$ bisects the perimeter and area of $\triangle ABC$, then $AS=6+x$, and
$$\begin{array}{l} 12=S_{\text {SSTT }}=\frac{1}{2}(6-x)(6+x) \sin A \\ =\frac{1}{2}\left(36-x^{2}\right) \cdot \frac{8}{10} . \end{array}$$

Thus, we have $x^{2}=6$. Solving for $x$ gives $x=\sqrt{6}$.
Hence, $AT=6-\sqrt{6}$, and $AS=6+\sqrt{6}$.
Therefore, the line $ST$ meets the requirements of the problem.
(3) As shown in Figure 2, when the line $l$ intersects $BC$ and $BA$ at points $S$ and $T$ respectively, if the line $l$ bisects the perimeter and area of $\triangle ABC$, let $BS=6-x$, then $BT=6+x$.
By the requirement of bisecting the area, we have
$$\begin{array}{l} 12=S_{\triangle B S T}=\frac{1}{2}(6-x)(6+x) \sin B \\ =\frac{1}{2}\left(36-x^{2}\right) \cdot \frac{3}{5} . \end{array}$$

Thus, we have $x^{2}=-4$.
Since the equation has no real solutions, such a line does not exist.
(4) As shown in Figure 3, when the line $l$ intersects $CA$ and $CB$ at points $S$ and $T$ respectively, if the line $l$ bisects the perimeter and area of $\triangle ABC$, let $CS=6-x$, then $CT$
$$\begin{aligned} =6+x & \text {, and still have } \\ 12 & =S_{\triangle \triangle T}=\frac{1}{2}\left(36-x^{2}\right) . \end{aligned}$$

Solving for $x$ gives $x=2 \sqrt{3}$.
At this point, $CS=6-2 \sqrt{3}$, and $CT=6+2 \sqrt{3}$.
However, $CT=6+2 \sqrt{3}>8=BC$, so such a line does not exist.

In summary, only the line in scenario (2) exists, meaning there is only one line that simultaneously bisects the perimeter and area of this triangle.