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If $2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}},-12 \hat{\mathbf{i}}-\hat{\mathbf{j}}-3 \hat{\mathbf{k}},-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$ and $\lambda \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ are the position vectors of four coplanar points, then $\lambda=$
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Verified Answer
The correct answer is:
6
If four points are collinear, then
$\begin{aligned}
\left|\begin{array}{lll}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
x_3-x_1 & y_3-y_1 & z_3-z_1 \\
x_4-x_1 & y_4-y_1 & z_4-z_1
\end{array}\right| & =0 \\
\left|\begin{array}{ccc}
-14 & 0 & -6 \\
-3 & 3 & -7 \\
\lambda-2 & 3 & -4
\end{array}\right| & =0
\end{aligned}$
So, $-14(-12+21)-0(-)-6(-9-3 \lambda+16)=0$
$\begin{aligned}
-126-6(-3-3 \lambda) & =0 \\
-126+18+18 \lambda & =0 \\
18 \lambda & =+108 \\
\lambda & =6
\end{aligned}$
$\begin{aligned}
\left|\begin{array}{lll}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
x_3-x_1 & y_3-y_1 & z_3-z_1 \\
x_4-x_1 & y_4-y_1 & z_4-z_1
\end{array}\right| & =0 \\
\left|\begin{array}{ccc}
-14 & 0 & -6 \\
-3 & 3 & -7 \\
\lambda-2 & 3 & -4
\end{array}\right| & =0
\end{aligned}$
So, $-14(-12+21)-0(-)-6(-9-3 \lambda+16)=0$
$\begin{aligned}
-126-6(-3-3 \lambda) & =0 \\
-126+18+18 \lambda & =0 \\
18 \lambda & =+108 \\
\lambda & =6
\end{aligned}$
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