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It $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are non-coplanar vectors and the four points with position vectors $2 \mathbf{a}+3 \mathbf{b}-\mathbf{c}$, $\mathbf{a}-2 \mathbf{b}+3 \mathbf{c}, 3 \mathbf{a}+4 \mathbf{b}+2 \mathbf{c}$ and $k \mathbf{a}-6 \mathbf{b}+6 \mathbf{c}$ are coplanar, then $k=$
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The correct answer is:
1
Let A(2a + 3b − c), B(a − 2b + 3c)
C(3a + 4b − 2c) and D(ka − 6b + 6c)
∴ AB = − a − 5b + 4c
$\begin{aligned}
& \mathbf{A C}=\mathbf{a}+\mathbf{b}-\mathbf{c} \\
& \mathbf{A D}=(k-2) \mathbf{a}-9 \mathbf{b}+7 \mathbf{c}
\end{aligned}$
Since, $A, B, C, D$ are coplanar
$\begin{array}{rlrl}
& & \left|\begin{array}{ccc}
-1 & -5 & 4 \\
1 & 1 & -1 \\
k-2 & -9 & 7
\end{array}\right|=0 \\
& \Rightarrow & (k-2) & (5-4)+9(1-4)+7(-1+5)=0 \\
& \Rightarrow & k-27+28 & =0 \\
& \Rightarrow & k-1 & =0 \\
& \Rightarrow & k & =1 .
\end{array}$
C(3a + 4b − 2c) and D(ka − 6b + 6c)
∴ AB = − a − 5b + 4c
$\begin{aligned}
& \mathbf{A C}=\mathbf{a}+\mathbf{b}-\mathbf{c} \\
& \mathbf{A D}=(k-2) \mathbf{a}-9 \mathbf{b}+7 \mathbf{c}
\end{aligned}$
Since, $A, B, C, D$ are coplanar
$\begin{array}{rlrl}
& & \left|\begin{array}{ccc}
-1 & -5 & 4 \\
1 & 1 & -1 \\
k-2 & -9 & 7
\end{array}\right|=0 \\
& \Rightarrow & (k-2) & (5-4)+9(1-4)+7(-1+5)=0 \\
& \Rightarrow & k-27+28 & =0 \\
& \Rightarrow & k-1 & =0 \\
& \Rightarrow & k & =1 .
\end{array}$
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