Step 1: Understanding the Concept:
For two lines to be perpendicular (at right angles), the dot product of their direction vectors must be zero. First, we must convert the given line equations into the standard symmetric form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ to correctly extract their direction ratios $\langle a, b, c \rangle$.
Step 2: Key Formula or Approach:
1. Standardize line equations: ensure the coefficients of $x, y, z$ in the numerators are exactly $+1$.
2. Perpendicularity condition: If line 1 has direction ratios $\langle a_1, b_1, c_1 \rangle$ and line 2 has $\langle a_2, b_2, c_2 \rangle$, then $a_1a_2 + b_1b_2 + c_1c_2 = 0$.
Step 3: Detailed Explanation:
Let's standardize the first line $L_1$:
\[ \frac{3-x}{2} = \frac{5y-2}{3\lambda+1} = 5 - z \]
Rewrite each term to have $+1$ coefficient for variables:
\[ \frac{-(x-3)}{2} = \frac{5(y - 2/5)}{3\lambda+1} = \frac{-(z-5)}{1} \]
\[ \frac{x-3}{-2} = \frac{y - 2/5}{(3\lambda+1)/5} = \frac{z-5}{-1} \]
The direction ratios for $L_1$ are $\vec{d}_1 = \langle -2, \frac{3\lambda+1}{5}, -1 \rangle$.
Let's standardize the second line $L_2$:
\[ \frac{x+2}{-1} = \frac{1-3y}{7} = \frac{4-z}{2\mu} \]
Rewrite:
\[ \frac{x+2}{-1} = \frac{-3(y - 1/3)}{7} = \frac{-(z-4)}{2\mu} \]
\[ \frac{x+2}{-1} = \frac{y - 1/3}{-7/3} = \frac{z-4}{-2\mu} \]
The direction ratios for $L_2$ are $\vec{d}_2 = \langle -1, -\frac{7}{3}, -2\mu \rangle$.
Since the lines are at right angles, their dot product is zero: $\vec{d}_1 \cdot \vec{d}_2 = 0$.
\[ (-2)(-1) + \left(\frac{3\lambda+1}{5}\right)\left(-\frac{7}{3}\right) + (-1)(-2\mu) = 0 \]
\[ 2 - \frac{7(3\lambda+1)}{15} + 2\mu = 0 \]
To clear the denominator, multiply the entire equation by 15:
\[ 30 - 7(3\lambda+1) + 30\mu = 0 \]
\[ 30 - 21\lambda - 7 + 30\mu = 0 \]
\[ 23 - 21\lambda + 30\mu = 0 \]
Rearranging terms to isolate $\lambda$ and $\mu$:
\[ 21\lambda - 30\mu = 23 \]
We need to find the value of $7\lambda - 10\mu$. Notice that $21\lambda - 30\mu$ is exactly 3 times the desired expression:
\[ 3(7\lambda - 10\mu) = 23 \]
Divide by 3:
\[ 7\lambda - 10\mu = \frac{23}{3} \]
Step 4: Final Answer:
The rigorously derived value is $23/3$.