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

1

The correct answer is Option (2) → 1

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

Write each numerator as $(\text{variable}-\text{point})$:

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

Direction ratios for line 1: $(l_1,m_1,n_1)=(5\lambda+2,\,-5,\,1)$

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

Rewrite:

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

Direction ratios for line 2: $(l_2,m_2,n_2)=(1,\,2\lambda,\,3)$

Perpendicular ⇒ $l_1l_2+m_1m_2+n_1n_2=0$

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

$5\lambda+2-10\lambda+3=0\;\Rightarrow\;-5\lambda+5=0\;\Rightarrow\;{\lambda=1}$