Let $L_1:\dfrac{x-1}{1}=\dfrac{y-2}{-1}=\dfrac{z-1}{2}$ and $L_2:\dfrac{x+1}{-1}=\dfrac{y}{2}=\dfrac{z}{1}$ be two lines. Let $L_3$ be a line passing through the point $(\alpha,\beta,\gamma)$ and be perpendicular to both $L_1$ and $L_2$. If $L_3$ intersects $L_1$, then $|5\alpha-11\beta-8\gamma|$ equals:
Step-by-Step Solution
Key Concept: The direction of $L_3$ is $\vec{m}\times\vec{n}$ where $\vec{m},\vec{n}$ are directions of $L_1,L_2$; then equate parametric forms of $L_3$ and $L_1$ to express $\alpha,\beta,\gamma$ in terms of parameters and evaluate the expression.
Direction of $L_3$: $\vec{m}\times\vec{n}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-1&2\\-1&2&1\end{vmatrix}=-5\hat{i}-3\hat{j}+\hat{k}$.
$L_3$: $\dfrac{x-\alpha}{-5}=\dfrac{y-\beta}{-3}=\dfrac{z-\gamma}{1}=\lambda$.
Point $A$ on $L_3$: $(\alpha-5\lambda,\beta-3\lambda,\gamma+\lambda)$. Point $B$ on $L_1$: $(k+1,-k+2,2k+1)$.
Equating: $\alpha=5\lambda+k+1$, $\beta=3\lambda-k+2$, $\gamma=-\lambda+2k+1$.
$|5\alpha-11\beta-8\gamma|=|5(5\lambda+k+1)-11(3\lambda-k+2)-8(-\lambda+2k+1)|=|-25|=25$.
Correct Answer: 3