Mathematics - Determinant Question with Solution | TestHub

Given,

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

$\Rightarrow \left(\begin{array}{ccc}2\alpha +9\beta +8\gamma & 10\alpha +3\beta +4\gamma & 8\alpha +8\beta +8\gamma \end{array}\right)=\left(\begin{array}{lll}0& 0& 0\end{array}\right)$

Now on comparing both side we get,

$2\alpha +2\beta +8\gamma =0 \dots \left(1\right)$

$10\alpha +3\beta +4\gamma =0 \dots \left(2\right)$

$8\alpha +8\beta +8\gamma =0 \dots \left(3\right)$

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

$\frac{\alpha }{1}=\frac{\beta }{6}=\frac{\gamma }{-7}=k\Rightarrow \alpha =k, \beta =6k, \gamma =-7k$

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

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

$\Rightarrow k=1$

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