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Let $\bar{a}=\lambda \bar{i}+3 \bar{j}+4 \bar{k}, \bar{b}=3 \bar{i}-\bar{j}+\lambda \bar{k}$ and $\bar{c}=\lambda \bar{i}+\bar{j}-3 \bar{k}$ be three vectors for some integer $\lambda$. If the volume of the parallelepiped with $\bar{a}, \bar{b}, \bar{c}$ as coterminus edges is 61 cubic units, then the number of possible values of $\lambda$ is
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The correct answer is:
$1$
Volume of parallelopiped $=\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]$
$\begin{aligned}
\Rightarrow & \left|\begin{array}{ccc}
\lambda & 3 & 4 \\
3 & -1 & \lambda \\
\lambda & 1 & -3
\end{array}\right|=61 \\
\Rightarrow & \lambda(3-\lambda)-3\left(-9-\lambda^2\right)+4(3+\lambda)=61 \\
\Rightarrow & 3 \lambda-\lambda^2+27+3 \lambda^2+12+4 \lambda=61 \\
\Rightarrow & 2 \lambda^2+7 \lambda-22=0 \\
\Rightarrow & 2 \lambda^2+11 \lambda-4 \lambda-22=0 \\
\Rightarrow & (2 \lambda+11)(\lambda-2)=0 \\
\Rightarrow & \lambda=\frac{-11}{2}, \lambda=2
\end{aligned}$
$\because \lambda$ is integer, $\therefore \lambda=2$.
Number of possible value is one.
$\begin{aligned}
\Rightarrow & \left|\begin{array}{ccc}
\lambda & 3 & 4 \\
3 & -1 & \lambda \\
\lambda & 1 & -3
\end{array}\right|=61 \\
\Rightarrow & \lambda(3-\lambda)-3\left(-9-\lambda^2\right)+4(3+\lambda)=61 \\
\Rightarrow & 3 \lambda-\lambda^2+27+3 \lambda^2+12+4 \lambda=61 \\
\Rightarrow & 2 \lambda^2+7 \lambda-22=0 \\
\Rightarrow & 2 \lambda^2+11 \lambda-4 \lambda-22=0 \\
\Rightarrow & (2 \lambda+11)(\lambda-2)=0 \\
\Rightarrow & \lambda=\frac{-11}{2}, \lambda=2
\end{aligned}$
$\because \lambda$ is integer, $\therefore \lambda=2$.
Number of possible value is one.
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