Let $\mathrm{L}_1: \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}$ and $\mathrm{L}_2: \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}$ 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…

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Let $\mathrm{L}_1: \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}$ and $\mathrm{L}_2: \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}$ 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 $\mathrm{L}_1$, then $|5 \alpha-11 \beta-8 \gamma|$ equals :

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { DR's of } L_3=\overrightarrow{\mathrm{m}} 	imes \overrightarrow{\mathrm{n}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \ 1 & -1 & 2 \ -1 & 2 & 1\end{array}ight| \ & =-5 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}} \ & L_3: \frac{x-\alpha}{-5}=\frac{y-\beta}{-3}=\frac{z-\gamma}{1}=\lambda \ & \quad A(\alpha-5 \lambda, \beta-3 \lambda, \gamma+\lambda) \ & L_1: \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}=k\end{aligned}$ $\mathrm{B}(\mathrm{k}+1,-\mathrm{k}+2,2 \mathrm{k}+1)$ Now $\begin{aligned} & \alpha-5 \lambda=\mathrm{k}+1 \Rightarrow \alpha=5 \lambda+\mathrm{k}+1 \ & \beta-3 \lambda=-\mathrm{k}+2 \Rightarrow \beta=3 \lambda-\mathrm{k}+2 \ & \gamma+\lambda=2 \mathrm{k}-1 \Rightarrow \gamma=-\lambda+2 \mathrm{k}+1 \ & |5 \alpha-11 \beta-8 \gamma|=|-25| \ & =25 \end{aligned}$ *

Let \(\mathrm{L}_1: \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}\) and \(\mathrm{L}_2:…

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