Question

If a point $P (\alpha, \beta, \gamma)$ satisfying
$(\alpha\,\, \beta\,\, \gamma) \begin{pmatrix} 2 & 10 & 8 \\9 & 3 & 8 \\8 & 4 & 8\end{pmatrix}=(0\,\,0\,\,0) $
lies on the plane $2 x+4 y+3 z=5$, then $6 \alpha+9 \beta+7 \gamma$ is equal to :

• A. $\frac{5}{4}$
• B. $-1$
• C. 11
• D. $\frac{11}{5}$

Answer 1

Answer: (C)

$ 2 \alpha+4 \beta+3 \gamma=5 $.........(1)
$ 2 \alpha+9 \beta+8 \gamma=0 $.........(2)
$10 \alpha+3 \beta+4 \gamma=0 $.........(3)
$ 8 \alpha+8 \beta+8 \gamma=0$.........(4)
Subtract (4) from (2)
$ -6 \alpha+\beta=0$
$ \beta=6 \alpha$.........(5)
From equation (4)
$ 8 \alpha+48 \alpha+8 \gamma=0 $
$ \gamma=-7 \alpha$.........(6)
From equation (1)
$ 2 \alpha+24 \alpha-21 \alpha=5$
$ 5 \alpha=5$
$ \alpha=1$
$ \beta=+6, \,\,\,\,\gamma=-7 $
$ \therefore 6 \alpha+9 \beta+7 \gamma $
$ =6+54-49$
$=11$