The vector equation of line passing through $(-1,3,-2)$ and perpendicular to the lines $\frac{x+4}{1}=\frac{y}{2}=\frac{z-3}{3}$ and $\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-6}{5}$ is

$\vec r=(-\hat i+3\hat j-2\hat k)+λ(2\hat i-7\hat j+4\hat k)$

The correct answer is Option (4) → $\vec r=(-\hat i+3\hat j-2\hat k)+λ(2\hat i-7\hat j+4\hat k)$

$\text{Given lines: }\frac{x+4}{1}=\frac{y}{2}=\frac{z-3}{3}\;\Rightarrow\;\vec d_1=\langle1,2,3\rangle$

$\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-6}{5}\;\Rightarrow\;\vec d_2=\langle-3,2,5\rangle$

$\text{Required direction }\vec d=\vec d_1\times\vec d_2= \begin{vmatrix} \hat i&\hat j&\hat k\\ 1&2&3\\ -3&2&5 \end{vmatrix} =4\hat i-14\hat j+8\hat k=2(2\hat i-7\hat j+4\hat k)$

$\text{Point on the line: }(-1,3,-2)\equiv-\hat i+3\hat j-2\hat k$

$\vec r=(-\hat i+3\hat j-2\hat k)+\lambda(2\hat i-7\hat j+4\hat k)$