Question:
If the point $(2, \alpha, \beta)$ lies on the plane which passes through the points $(3,4,2)$ and $(7,0,6)$ and is perpendicular to the plane $2 x-5 y=15$, then $2 \alpha$ $-3 \beta$ is equal to :-
Correct Option: , 4
Solution:
Normal vector of plane
$=\left|\begin{array}{ccc}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 2 & -5 & 0 \\ 4 & -4 & 4\end{array}\right|=-4(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})$
equation of plane is $5(x-7)+2 y-3(z-6)=0$ $5 x+2 y-3 z=17$
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