If lines $\frac{x+5}{5\lambda+2} = \frac{4-2y}{10} = \frac{1-3z}{-3}$ and $\frac{x-2}{1} = \frac{1+2y}{4\lambda} = \frac{2+z}{3}$ are perpendicular, than value of '$\lambda$' is

We need to find the value of $\lambda$ that makes the two given lines perpendicular to each other.

---

First, let's rewrite the lines in standard form so we can read off the direction ratios.

For Line 1: $\dfrac{x+5}{5\lambda+2} = \dfrac{4-2y}{10} = \dfrac{1-3z}{-3}$

The $y$-part:

$\dfrac{4 - 2y}{10} = \dfrac{-(2y - 4)}{10} = \dfrac{y - 2}{-5}$

The $z$-part:

$\dfrac{1 - 3z}{-3} = \dfrac{-(3z - 1)}{-3} = \dfrac{z - \frac{1}{3}}{1}$

So Line 1 becomes:

$\dfrac{x+5}{5\lambda+2} = \dfrac{y-2}{-5} = \dfrac{z - \frac{1}{3}}{1}$

Direction ratios of Line 1: $(5\lambda + 2,\ -5,\ 1)$

---

For Line 2: $\dfrac{x-2}{1} = \dfrac{1+2y}{4\lambda} = \dfrac{2+z}{3}$

The $y$-part:

$\dfrac{1 + 2y}{4\lambda} = \dfrac{2(y + \frac{1}{2})}{4\lambda} = \dfrac{y + \frac{1}{2}}{2\lambda}$

So Line 2 becomes:

$\dfrac{x-2}{1} = \dfrac{y + \frac{1}{2}}{2\lambda} = \dfrac{z+2}{3}$

Direction ratios of Line 2: $(1,\ 2\lambda,\ 3)$

---

Two lines are perpendicular when the dot product of their direction ratios equals zero:

$a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$

Substituting:

$(5\lambda + 2)(1) + (-5)(2\lambda) + (1)(3) = 0$

$5\lambda + 2 - 10\lambda + 3 = 0$

$-5\lambda + 5 = 0$

$-5\lambda = -5$

$\lambda = 1$

---

Verification with $\lambda = 1$:

Line 1 direction ratios: $(7,\ -5,\ 1)$ and Line 2 direction ratios: $(1,\ 2,\ 3)$

$(7)(1) + (-5)(2) + (1)(3) = 7 - 10 + 3 = 0$ ✅

---

$\boxed{\lambda = 1}$