If a point P(,,) satisfying [( )pmatrix2&10&89&3&88&4&8pmatrix=(0 0 0)] lies on the plane …

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

Accepted Answer

The matrix equation gives:
$$2\alpha+9\beta+8\gamma=0,\qquad 10\alpha+3\beta+4\gamma=0,\qquad 8\alpha+8\beta+8\gamma=0,$$
together with the plane condition $2\alpha+4\beta+3\gamma=5$.

Subtracting the third from the first: $-6\alpha+\beta=0\Rightarrow\beta=6\alpha$.
From the third: $8\alpha+48\alpha+8\gamma=0\Rightarrow\gamma=-7\alpha$.
Substituting into the plane equation: $2\alpha+24\alpha-21\alpha=5\Rightarrow5\alpha=5\Rightarrow\alpha=1$.
$$\beta=6,\qquad \gamma=-7.$$
$$6\alpha+9\beta+7\gamma=6+54-49=11.$$

Solutions · 1