If a point P(α ,β ,γ ) satisfying ( … ) lies on the plane 2x+4y+3z=5, then 6α +9β +7γ is equal to

Given,

$$(\begin{matrix}\alpha & \beta & \gamma \end{matrix})(\begin{matrix}2 & 10 & 8 \\ 9 & 3 & 8 \\ 8 & 4 & 8\end{matrix})=(\begin{matrix}0 & 0 & 0\end{matrix})$$(α​β​γ​)(298​1034​888​)=(0​0​0​)

$$\Rightarrow (\begin{matrix}2\alpha +9\beta +8\gamma & 10\alpha +3\beta +4\gamma & 8\alpha +8\beta +8\gamma \end{matrix})=(\begin{matrix}0 & 0 & 0\end{matrix})$$⇒(2α+9β+8γ​10α+3β+4γ​8α+8β+8γ​)=(0​0​0​)

Now on comparing both side we get,

$$2\alpha +2\beta +8\gamma =0\ldots (1)$$2α+2β+8γ=0…(1)

$$10\alpha +3\beta +4\gamma =0\ldots (2)$$10α+3β+4γ=0…(2)

$$8\alpha +8\beta +8\gamma =0\ldots (3)$$8α+8β+8γ=0…(3)

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

$$\frac{\alpha }{1}=\frac{\beta }{6}=\frac{\gamma }{-7}=k\Rightarrow \alpha =k,\beta =6k,\gamma =-7k$$1α​=6β​=−7γ​=k⇒α=k,β=6k,γ=−7k

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

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

$$\Rightarrow k=1$$⇒k=1

Hence, $6\alpha +9\beta +7\gamma =6k+54k-49k=11k=11$6α+9β+7γ=6k+54k−49k=11k=11.