If the line $$\frac{2-x}{3} = \frac{3y-2}{4\lambda+1} = 4-z$$ makes a right angle with the line $$\frac{x+3}{3\mu} = \frac{1-2y}{6} = \frac{5-z}{7}$$, then $$4\lambda + 9\mu$$ is equal to :

Standardize the direction ratios (DRs):

$$L_1: \frac{x-2}{-3} = \frac{y-2/3}{(4\lambda+1)/3} = \frac{z-4}{-1}$$. DRs: $$\vec{v_1} = (-3, \frac{4\lambda+1}{3}, -1)$$.

$$L_2: \frac{x+3}{3\mu} = \frac{y-1/2}{-3} = \frac{z-5}{-7}$$. DRs: $$\vec{v_2} = (3\mu, -3, -7)$$.

For perpendicular lines, $$\vec{v_1} \cdot \vec{v_2} = 0$$:

$$(-3)(3\mu) + (\frac{4\lambda+1}{3})(-3) + (-1)(-7) = 0$$

$$-9\mu - (4\lambda + 1) + 7 = 0$$

$$-9\mu - 4\lambda + 6 = 0 \implies \mathbf{4\lambda + 9\mu = 6}$$