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A line with direction cosines proportional to 2, 1, 2 meets the line $L_1$ passing through $(0,-1,0)$ with direction ratios $1,1,1$ at $A(x, y, z)$ and another line $L_2$ at $B(1,1,1)$ then $x+y+z=$
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The correct answer is:
8
Equation of line $L_1$ passing through $(0,-1,0)$
and with direction ratio $(1,1,1)$ is
$\begin{aligned}
\qquad \frac{x}{1}=\frac{y+1}{1}=\frac{z}{1} \\
\therefore \quad A(x, y, z) \equiv(\lambda, \lambda-1, \lambda) \\
\text { A line intersect } A(x, y, z) \text { and } B(1,1,1) \\
\therefore \text { direction ratio of } A B \\
\quad(\lambda-1, \lambda-2, \lambda-1)
\end{aligned}$
Given direction ratio of line intersect $A B$ is 2, 1, 2
$\begin{aligned}
& \therefore \quad \lambda-1=2, \lambda-2=1, \lambda-1=2 \Rightarrow \lambda=3 \\
& \because \quad x=3, y=2, z=3 \\
& x+y+z=3+2+3=8
\end{aligned}$
and with direction ratio $(1,1,1)$ is
$\begin{aligned}
\qquad \frac{x}{1}=\frac{y+1}{1}=\frac{z}{1} \\
\therefore \quad A(x, y, z) \equiv(\lambda, \lambda-1, \lambda) \\
\text { A line intersect } A(x, y, z) \text { and } B(1,1,1) \\
\therefore \text { direction ratio of } A B \\
\quad(\lambda-1, \lambda-2, \lambda-1)
\end{aligned}$
Given direction ratio of line intersect $A B$ is 2, 1, 2
$\begin{aligned}
& \therefore \quad \lambda-1=2, \lambda-2=1, \lambda-1=2 \Rightarrow \lambda=3 \\
& \because \quad x=3, y=2, z=3 \\
& x+y+z=3+2+3=8
\end{aligned}$
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