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

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Aicharya · Accepted Answer

Given, $\begin{bmatrix} \alpha & \beta & \gamma \end{bmatrix} \begin{bmatrix} 2 & 10 & 8 \\ 9 & 3 & 8 \\ 8 & 4 & 8 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}$ $\Rightarrow \begin{bmatrix} 2\alpha + 9\beta + 8\gamma & 10\alpha + 3\beta + 4\gamma & 8\alpha + 8\beta + 8\gamma \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}$

Now on comparing both side we get,

$2 α + 2 β + 8 γ = 0 \dots (1)$

$10 α + 3 β + 4 γ = 0 \dots (2)$

$8 α + 8 β + 8 γ = 0 \dots (3)$

Now from $(1)$ and $(3)$ by cross multiplication we get,

$\frac{α}{1} = \frac{β}{6} = \frac{γ}{- 7} = k \Rightarrow α = k, β = 6 k, γ = - 7 k$

Point $P (α, β, γ)$ lie on the plane $2 x + 4 y + 3 z = 5 \Rightarrow 2 α + 4 β + 3 γ = 5$

$\Rightarrow 2 k + 24 k - 21 k = 5$

$\Rightarrow k = 1$

Hence, $6 α + 9 β + 7 γ = 6 k + 54 k - 49 k = 11 k = 11$ .

If a point $P(\alpha, \beta, \gamma)$ satisfying $\begin{bmatrix} \alpha & \beta & \gamma \end{bmatrix}…

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