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)

11

Answer 2

2α+4β+3γ=5.........(1)
2α+9β+8γ=0.........(2)
10α+3β+4γ=0.........(3)
8α+8β+8γ=0.........(4)
Subtract (4) from (2)
−6α+β=0
β=6α.........(5)
From equation (4)
8α+48α+8γ=0
γ=−7α.........(6)
From equation (1)
2α+24α−21α=5
5α=5
α=1
β=+6,γ=−7
∴6α+9β+7γ
=6+54−49
=11
So, the correct option is (C) : 11