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The point of intersection of the lines $\frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1}, \frac{x+3}{-36}=\frac{y-3}{2}=\frac{z-6}{4}$ is
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Verified Answer
The correct answer is:
$21, \frac{5}{3}, \frac{10}{3}$
Given lines are,
$\begin{aligned}
& \qquad \frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1}=r_1, \text { (say) } \\
& \text { and } \frac{x+3}{-36}=\frac{y-3}{2}=\frac{z-6}{4}=r_2, \text { (say) } \\
& \therefore x=3 r_1+5=-36 r_2-3, y=-r_1+7=3+2 r_2 \text { and } z=r_1-2=4 r_2+6 \\
& \text { On solving, we get } x=21, y=\frac{5}{3}, z=\frac{10}{3} .
\end{aligned}$
On solving, we get
Trick: Check through options.
$\begin{aligned}
& \qquad \frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1}=r_1, \text { (say) } \\
& \text { and } \frac{x+3}{-36}=\frac{y-3}{2}=\frac{z-6}{4}=r_2, \text { (say) } \\
& \therefore x=3 r_1+5=-36 r_2-3, y=-r_1+7=3+2 r_2 \text { and } z=r_1-2=4 r_2+6 \\
& \text { On solving, we get } x=21, y=\frac{5}{3}, z=\frac{10}{3} .
\end{aligned}$
On solving, we get
Trick: Check through options.
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