It is given that the lines
$\mathbf{r}=\hat{\mathbf{i}}-6 \hat{\mathbf{j}}(p \sec \alpha) \hat{\mathbf{k}}+t(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})$
and $\mathbf{r}=4 \hat{\mathbf{j}}+\hat{\mathbf{k}}+\lambda(2 \hat{\mathbf{i}}+(p \tan \alpha) \hat{\mathbf{j}}+2 \hat{\mathbf{k}})$
Intersected at point $8 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}+9 \hat{\mathbf{k}}$, so $1+t=8,-6+2 t=8$ and $p \sec \alpha+t=9$
$\therefore \quad t=7$ and $p \sec \alpha=2$ $\ldots(\mathrm{i})$
and $2 \lambda=8,4+\lambda p \tan \alpha=8$ and $1+2 \lambda=9$
$\therefore \quad \lambda=4$ and $p \tan \alpha=1$ $\ldots(\mathrm{ii})$
From Eqs. (i) and (ii), we get
$p^2 \sec ^2 \alpha-p^2 \tan ^2 \alpha=4-1 \Rightarrow p^2=3 \Rightarrow p= \pm \sqrt{3}$