\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+17}{p}\)
\(\overrightarrow{b_1}=2 \hat{i}+3 \hat{j}+p \hat{k} \text { and } \frac{x+4}{2}=\frac{y+9}{2}=\frac{z-1}{2}\)
\(\overrightarrow{b_{2}}=2 \hat{i}+2 \hat{j}+2 \hat{k}\)
Since the lines are perpendicular.
Hence \(\overrightarrow{b_{1}} \cdot \vec{b}_{2}=0 \)
\(\Rightarrow 4+6+2 p=0 \)
\(\Rightarrow 2 p=-10 \)
\(\Rightarrow p=-5\)
