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(B) The lines intersect at the point $8\hat{i} + 8\hat{j} + 9\hat{k}$.For the first line: $r = (1+t)\hat{i} + (-6+2t)\hat{j} + (p \sec \alpha + t)\hat

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$\sqrt{3}$

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(B) The lines intersect at the point $8\hat{i} + 8\hat{j} + 9\hat{k}$.For the first line: $r = (1+t)\hat{i} + (-6+2t)\hat{j} + (p \sec \alpha + t)\hat{k}$.Equating components with $8\hat{i} + 8\hat{j} + 9\hat{k}$:$1+t = 8 \Rightarrow t = 7$.$-6+2t = 8 \Rightarrow -6+14 = 8$ (consistent).$p \sec \alpha + t = 9 \Rightarrow p \sec \alpha + 7 = 9 \Rightarrow p \sec \alpha = 2$ ... $(i)$.For the second line: $r = (2\lambda)\hat{i} + (4 + \lambda p 	an \alpha)\hat{j} + (1 + 2\lambda)\hat{k}$.Equating components with $8\hat{i} + 8\hat{j} + 9\hat{k}$:$2\lambda = 8 \Rightarrow \lambda = 4$.$1 + 2\lambda = 9 \Rightarrow 1 + 8 = 9$ (consistent).$4 + \lambda p 	an \alpha = 8 \Rightarrow 4 + 4p 	an \alpha = 8 \Rightarrow 4p 	an \alpha = 4 \Rightarrow p 	an \alpha = 1$ ... $(ii)$.Squaring and subtracting $(ii)$ from $(i)$:$(p \sec \alpha)^2 - (p 	an \alpha)^2 = 2^2 - 1^2$.$p^2(\sec^2 \alpha - 	an^2 \alpha) = 4 - 1$.Since $\sec^2 \alpha - 	an^2 \alpha = 1$, we have $p^2 = 3$.Given $0 < \alpha < \frac{\pi}{2}$, $p$ must be positive, so $p = \sqrt{3}$.

If the point of intersection of the lines $r = \hat{i} - 6\hat{j} + (p \sec \alpha) \hat{k} + t(\hat{i} + 2\hat{j} + \hat{k})$ and $r = 4\hat{j} + \hat{k} + \lambda(2\hat{i} + (p \tan \alpha) \hat{j} + 2\hat{k})$ is $8\hat{i} + 8\hat{j} + 9\hat{k}$, (where $0 < \alpha < \frac{\pi}{2}$), then $p =$

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