Three Dimensional Geometry Question 28
Question 28 - 31 January - Shift 2
If a point $P(\alpha, \beta, \gamma)$ satisfying $(\alpha ; \beta ; \gamma)\left(\begin{array}{ccc}2 & 10 & 8 \\ 9 & 3 & 8 \\ 8 & 4 & 8\end{array}\right)=\left(\begin{array}{lll}0 & 0 & 0\end{array}\right)$ lies on the plane $2 x+4 y+3 z=5$, then $6 \alpha+9 \beta+7 \gamma$ is equal to :
(1) -1
(2) $\frac{11}{5}$
(3) $\frac{5}{4}$
(4) 11
Show Answer
Answer: (4)
Solution:
Formula: Equation Of A Plane
$2 \alpha+4 \beta+3 \gamma=5 \ldots \ldots (1)$
$2 \alpha+9 \beta+8 \gamma=0 \ldots \ldots (2)$
$10 \alpha+3 \beta+4 \gamma=0 \ldots \ldots (3)$
$8 \alpha+8 \beta+8 \gamma=0 \ldots \ldots (4)$
Subtract (4) from (2)
$-6 \alpha+\beta=0$
$\beta=6 \alpha$
From equation (4)
$8 \alpha+48 \alpha+8 \gamma=0$
$\gamma=-7 \alpha$
From equation (1)
$2 \alpha+24 \alpha-21 \alpha=5$
$5 \alpha=5$
$\alpha=1$
$\beta=+6, \quad \gamma=-7$
$\therefore 6 \alpha+9 \beta+7 \gamma$
$=6+54-49$
$=11$
