Question

If the point $(2,\alpha ,\beta )$ lies on the plane which passes through the point $(3,4,2)$ and$(7,0,6)$ and is perpendicular to the plane $2x-5y=15$, then $2\alpha -3\beta$ is equal to :

• A. $17$
• B. $12$
• C. $5$
• D. $7$

Answer 1

The correct option is D $7$

Parallel vector to the plane $P_1$ is $(4,-4,4)\equiv (1,-1,1)$
Normal vector to the plane $P_2:2x-5y=15$ is $(2,-5,0)$

$\therefore$ Normal vector to the plane $P_1$ is
$\overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 1\\ 2& -5& 0\end{array}\right|=5\hat{i}+2\hat{j}-3\hat{k}$

So, the plane $P_1$ with normal vector $\overrightarrow{n}=5\hat{i}+2\hat{j}-3\hat{k}$ passing through $(3,4,2)$ is given by
$5(x-3)+2(y-4)-3(z-2)=0$
$\Rightarrow 5x+2y-3z-17=0$

$(2,\alpha ,\beta )$ lies on the plane $P_1$
$\therefore 5\times 2+2\alpha -3\beta -17=0\Rightarrow 2\alpha -3\beta =7$