If a point P(, , ) satisfying [( pmatrix & & pmatrix )( pmatrix 2 & 10 & 8 9 & 3 & 8 8 & 4…

Question

If a point $\mathrm{P}(\alpha, \beta, \gamma)$ satisfying

$$\left( {\begin{pmatrix} \alpha  & \beta  & \gamma   \ \end{pmatrix} } \right)\left( {\begin{pmatrix} 2 & {10} & 8  \
   9 & 3 & 8  \
   8 & 4 & 8  \ \end{pmatrix} } \right) = \left( {\begin{pmatrix} 0 & 0 & 0  \ \end{pmatrix} } \right)$$

lies on the plane $2 x+4 y+3 z=5$, then $6 \alpha+9 \beta+7 \gamma$ is equal to :

Accepted Answer

Point $\mathrm{P}(\alpha, \beta, \gamma)$ lies on the plane $2 x+4 y+3 z=5$,

$$ \therefore $$ $2 \alpha+4 \beta+3 \gamma=5$ ........(1)

Given, $$\left( {\begin{pmatrix} \alpha  & \beta  & \gamma   \

\end{pmatrix} } \right)\left( {\begin{pmatrix} 2 & {10} & 8  \
   9 & 3 & 8  \
   8 & 4 & 8  \

\end{pmatrix} } \right) = \left( {\begin{pmatrix} 0 & 0 & 0  \

\end{pmatrix} } \right)$$

$$ \therefore $$ $2 \alpha+9 \beta+8 \gamma=0$ .......(2)

and $10 \alpha+3 \beta+4 \gamma=0$ ........(3)

and $8 \alpha+8 \beta+8 \gamma=0$ ..........(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)

$2 \alpha+24 \alpha-21 \alpha=5$

$5 \alpha=5$

$\alpha=1$

$\beta=+6, \quad \gamma=-7$

$\therefore 6 \alpha+9 \beta+7 \gamma$

$=6+54-49$

$=11$

Solutions · 1