The value of P so that the lines $\frac{1-\mathrm{x}}{3}=\frac{7\mathrm{y}-14}{2\mathrm{p}}=\frac{\mathrm{z}-3}{2}\text{ and }\frac{7-7\mathrm{x}}{3\mathrm{p}}=\frac{\mathrm{y}-5}{1}=\frac{6-\mathrm{z}}{5}$ are at right angle is _______ (where [.] = GIF)

$\begin{array}{l}\frac{\mathrm{x}-1}{-3}=\frac{\mathrm{y}-2}{\frac{2\mathrm{p}}{7}}=\frac{\mathrm{z}-3}{2}---\left(1\right)\\ \Rightarrow \frac{\mathrm{x}-1}{{\displaystyle \frac{-3p}{7}}}=\frac{\mathrm{y}-5}{1}=\frac{\mathrm{z}-6}{-5}---\left(2\right)\end{array}$

(1) and (2) are perpendicular.

$$\begin{gathered} \Rightarrow {\mathrm{a}}_1{\mathrm{a}}_2+{\mathrm{b}}_1{\mathrm{b}}_2+{\mathrm{c}}_1{\mathrm{c}}_2=0 \\ \Rightarrow \left(-3\right)\left(\frac{-3p}{7}\right)+\left(\frac{2p}{7}\right)1+\left(2\right)\left(-5\right)=0 \\ \Rightarrow \frac{9p}{7}+\frac{2p}{7}-10=0\Rightarrow 11p=70 \\ \Rightarrow \left[p\right]=\left[\frac{70}{11}\right]=6 \end{gathered}$$