Let $A$ be the point of intersection of the lines $$L_1:\;\frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1},$$ and $$L_2:\;\frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}.$$ Let $B$ and $C$ be points on $L_1$ and $L_2$ respectively such that $AB=AC=\sqrt{15}$. Then the square of the area of triangle $ABC$ is:

Step 1: Find the intersection point A

First, we express the parametric equations for both lines $L_1$ and $L_2$. For $L_1$, we set $\frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1}=t$, which gives $x=7+t$, $y=5$, and $z=3-t$. For $L_2$, we set $\frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}=s$, which gives $x=1+3s$, $y=-3+4s$, and $z=-7+5s$. To find the intersection point $A$, we equate the corresponding coordinates.

Step 2: Solve for parameters and point A

Equating the $y$ -coordinates, we get $5 = -3+4s$, which simplifies to $4s=8$, so $s=2$. Substitute $s=2$ into the equations for $x$ and $z$. From $7+t = 1+3s$, we get $7+t = 1+3(2) = 7$, so $t=0$. From $3-t = -7+5s$, we get $3-t = -7+5(2) = 3$, so $t=0$. Both equations yield $t=0$. Thus, the intersection point $A$ is found by substituting $t=0$ into $L_1$ 's equations or $s=2$ into $L_2$ 's equations.

Step 3: Calculate coordinates of A

Using $t=0$ in the parametric equations for $L_1$, we find the coordinates of point $A$: $x = 7+0=7$, $y=5$, $z=3-0=3$. So, $A=(7{,}5{,}3)$.

Step 4: Find direction vectors of L1 and L2

The direction vector for $L_1$ is $\vec{d}_{1} = \langle 1, 0, -1 \rangle$ and for $L_2$ is $\vec{d}_{2} = \langle 3, 4, 5 \rangle$. These vectors represent the directions along which points $B$ and $C$ lie from $A$ respectively.

Step 5: Calculate vectors AB and AC

Points $B$ and $C$ are on $L_1$ and $L_2$ respectively, such that $AB=AC=\sqrt{15}$. We can find the vectors $\vec{AB}$ and $\vec{AC}$ by scaling the direction vectors $\vec{d}_{1}$ and $\vec{d}_{2}$ by the given length. The magnitude of $\vec{d}_{1}$ is $\sqrt{1^2+0^2+(-1)^2} = \sqrt{2}$. The magnitude of $\vec{d}_{2}$ is $\sqrt{3^2+4^2+5^2} = \sqrt{9+16+25} = \sqrt{50}$.

Step 6: Calculate the square of the area of triangle ABC

The square of the area of triangle $ABC$ can be calculated using the formula $\text{Area}^2 = \frac{1}{4} |\vec{AB} \times \vec{AC}|^2$. First, we compute the cross product of $\vec{AB}$ and $\vec{AC}$. We can choose the positive signs for the scaled vectors. The cross product $\langle 1, 0, -1 \rangle \times \langle 3, 4, 5 \rangle = \langle (0)(5)-(-1)(4), (-1)(3)-(1)(5), (1)(4)-(0)(3) \rangle = \langle 4, -8, 4 \rangle$. Then, we multiply by the scalar factor $\frac{15}{\sqrt{2}\sqrt{50}} = \frac{15}{\sqrt{100}} = \frac{15}{10} = \frac{3}{2}$. So, $\vec{AB} \times \vec{AC} = \frac{3}{2} \langle 4, -8, 4 \rangle = \langle 6, -12, 6 \rangle$. Finally, we calculate the magnitude squared of this vector and divide by 4.

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