Question

Let $\alpha ,\beta ,\gamma $ be three real numbers satisfying $\left[\alpha \, \, \beta \, \, \gamma \right]\begin{bmatrix} \, \, 2 & -1 & \, \, 1 \\ -1 & -1 & -2 \\ -1 & \, \, 2 & \, \, 1 \end{bmatrix}=\left[0 \, \, 0 \, \, 0\right]$ . If the point $A\left(\alpha , \beta , \gamma \right)$ lies on the plane $2x+y+3z=2$ , then $3\alpha +2\beta -6\gamma $ is equal to

• A. $0$
• B. $-\frac{1}{3}$
• C. $1$
• D. $-3$

Answer 1

Answer: (B)

$\left[\alpha \, \beta \, \gamma \right]\begin{bmatrix} 2 & -1 & 1 \\ -1 & -1 & -2 \\ -1 & 2 & 1 \end{bmatrix}=\left[0 \, 0 \, 0\right]$
$\begin{bmatrix} 2\alpha -\beta -\gamma & -\alpha -\beta +2\gamma & \alpha -2\beta +\gamma \end{bmatrix}=\left[0 \, 0 \, 0\right]$
$\Rightarrow 2\alpha -\beta -\gamma =0$
$-\alpha -\beta +2\gamma =0$
$\alpha -2\beta +\gamma =0$
Using cramer's rule $\Delta=\left|\begin{array}{ccc}2 & -1 & -1 \\ -1 & -1 & 2 \\ 1 & -2 & 1\end{array}\right|=2(-1+4)+1(-1-2)-1(2+1)$ $=6-3-3=0$ Using cramer's rule
$\Delta=\left|\begin{array}{ccc}2 & -1 & -1 \\ -1 & -1 & 2 \\ 1 & -2 & 1\end{array}\right|=2(-1+4)+1(-1-2)-1(2+1)$
$=6-3-3=0$
So, the system of equations has infinite solutions Put $\gamma=k$
$\left.\begin{array}{l}2 \alpha-\beta=k \\ \alpha+\beta=2 k\end{array}\right\} \Rightarrow \gamma=k, \beta=k$
$\Rightarrow (\alpha, \beta, \gamma) \equiv(k, k, k)$
$\because(\alpha, \beta, \gamma)$ satisfies $2 x+y+3 z=2$
$\Rightarrow 6 k=2 \Rightarrow k=\frac{1}{3}$
$(\alpha, \beta, \gamma) \equiv\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)$
$\Rightarrow 3 \alpha+2 \beta-6 \gamma=-\frac{1}{3}$