1. Write the given lines in parametric form

The first line is $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda$ so its parametric equations are $x=1+2\lambda,\quad y=-1+3\lambda,\quad z=1+4\lambda.$ Hence a general point on this line is $P(1+2\lambda,-1+3\lambda, 1+4\lambda).$

The second line is $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z}{1}=\mu$ so its parametric equations are $x=3+\mu,\quad y=4+2\mu,\quad z=\mu.$ Hence a general point on this line is $Q(3+\mu, 4+2\mu, \mu).$

2. Use the fact that line $L$ passes through $(1,1,1)$ and intersects both given lines

So there exist parameters $\lambda,\mu$ such that the three points $A=(1,1,1),\quad P=(1+2\lambda,-1+3\lambda,1+4\lambda),\quad Q=(3+\mu,4+2\mu,\mu)$ are collinear.

That means vectors $\overrightarrow{AP}$ and $\overrightarrow{AQ}$ are parallel.

Compute: $\overrightarrow{AP}=P-A=(2\lambda, 3\lambda-2, 4\lambda)$ and $\overrightarrow{AQ}=Q-A=(2+\mu, 3+2\mu, \mu-1).$

For collinearity, $\overrightarrow{AP}=k\,\overrightarrow{AQ}$ for some scalar $k$.

So, $2\lambda=k(2+\mu) \quad ...(1)$ $3\lambda-2=k(3+2\mu) \quad ...(2)$ $4\lambda=k(\mu-1) \quad ...(3)$

3. Eliminate $k$ and solve for $\lambda,\mu$

From (1) and (3), $\frac{2\lambda}{2+\mu}=\frac{4\lambda}{\mu-1}$ Assuming $\lambda\neq 0$ (otherwise $P=A$, which does not lie on the first line), cancel $\lambda$: $\frac{2}{2+\mu}=\frac{4}{\mu-1}$ $2(\mu-1)=4(2+\mu)$ $2\mu-2=8+4\mu$ $-10=2\mu$ $\mu=-5.$

Now from (1), $2\lambda=k(2-5)=k(-3)$ so $k=-\frac{2\lambda}{3}.$

Use (2): $3\lambda-2=k(3+2(-5))=k(-7).$ Substitute $k=-\frac{2\lambda}{3}$: $3\lambda-2=\left(-\frac{2\lambda}{3}\right)(-7)=\frac{14\lambda}{3}.$ Multiply by $3$: $9\lambda-6=14\lambda$ $-6=5\lambda$ $\lambda=-\frac{6}{5}.$

Thus, $\mu=-5,\qquad \lambda=-\frac{6}{5}.$

4. Find the two intersection points

Point on first line: $P=\left(1+2\left(-\frac65\right), -1+3\left(-\frac65\right), 1+4\left(-\frac65\right)\right)$ $P=\left(1-\frac{12}{5}, -1-\frac{18}{5}, 1-\frac{24}{5}\right)$ $P=\left(-\frac75, -\frac{23}{5}, -\frac{19}{5}\right).$

Point on second line: $Q=(3-5, 4+2(-5), -5)=(-2,-6,-5).$

The required line $L$ passes through $A=(1,1,1)$ and also through $Q=(-2,-6,-5)$. So a direction vector of $L$ is $\overrightarrow{AQ}=(-3,-7,-6).$ Hence equation of $L$ can be taken as $\frac{x-1}{-3}=\frac{y-1}{-7}=\frac{z-1}{-6}.$ Or parametric form: $x=1-3t,\quad y=1-7t,\quad z=1-6t.$

5. Check the options

We test whether each point satisfies $x=1-3t,\quad y=1-7t,\quad z=1-6t.$

Option A: $(7,15,13)$

From $x=1-3t$, $7=1-3t \Rightarrow t=-2.$ Then $y=1-7(-2)=15,\quad z=1-6(-2)=13.$ Both match.

So A lies on $L$.

Option B: $(4,22,7)$

From $x=4$, $4=1-3t \Rightarrow t=-1.$ Then $y=1-7(-1)=8\neq 22.$ So B does not lie on $L$.

Option C: $(10,-29,-50)$

From $x=10$, $10=1-3t \Rightarrow t=-3.$ Then $y=1-7(-3)=22\neq -29.$ So C does not lie on $L$.

Option D: $(5,4,3)$

From $x=5$, $5=1-3t \Rightarrow t=-\frac43.$ Then $y=1-7\left(-\frac43\right)=\frac{31}{3}\neq 4.$ So D does not lie on $L$.

6. Conclusion

The point lying on line $L$ is $\boxed{(7,15,13)}$ which is Option A.