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If the lines $\frac{x-\mathrm{k}}{2}=\frac{y+1}{3}=\frac{\mathrm{z}-1}{4} \quad$ and $\frac{x-3}{1}=\frac{y-\frac{9}{2}}{2}=\frac{\mathrm{z}}{1}$ intersect, then the value of $\mathrm{k}$ is
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Verified Answer
The correct answer is:
$1$
Since the given lines intersect
$\begin{aligned}
&\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & \mathrm{z}_2-\mathrm{z}_1 \\
\mathrm{a}_1 & \mathrm{~b}_1 & \mathrm{c}_1 \\
\mathrm{a}_2 & \mathrm{~b}_2 & \mathrm{c}_2
\end{array}\right|=0 \\
& \therefore \quad\left|\begin{array}{ccc}
3-\mathrm{k} & \frac{9}{2}+1 & 0-1 \\
2 & 3 & 4 \\
1 & 2 & 1
\end{array}\right|=0 \\
& \Rightarrow\left|\begin{array}{ccc}
3-\mathrm{k} & \frac{11}{2} & -1 \\
2 & 3 & 4 \\
1 & 2 & 1
\end{array}\right|=0 \\
& \Rightarrow-5+5 \mathrm{k}=0 \\
& \Rightarrow \mathrm{k}=1
\end{aligned}$
$\begin{aligned}
&\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & \mathrm{z}_2-\mathrm{z}_1 \\
\mathrm{a}_1 & \mathrm{~b}_1 & \mathrm{c}_1 \\
\mathrm{a}_2 & \mathrm{~b}_2 & \mathrm{c}_2
\end{array}\right|=0 \\
& \therefore \quad\left|\begin{array}{ccc}
3-\mathrm{k} & \frac{9}{2}+1 & 0-1 \\
2 & 3 & 4 \\
1 & 2 & 1
\end{array}\right|=0 \\
& \Rightarrow\left|\begin{array}{ccc}
3-\mathrm{k} & \frac{11}{2} & -1 \\
2 & 3 & 4 \\
1 & 2 & 1
\end{array}\right|=0 \\
& \Rightarrow-5+5 \mathrm{k}=0 \\
& \Rightarrow \mathrm{k}=1
\end{aligned}$
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