Question

Let the equation of the plane, that passes through the point $(1,4,-3)$ and contains the line of intersection of the planes $3x-2y+4z-7=0$ and $x+5y-2z+9=0,$ be $\alpha x+\beta y+\gamma z+3=0,$ then $\alpha +\beta +\gamma$ is equal to

• A. $-23$
• B. $-15$
• C. $23$
• D. $15$

Answer 1

The correct option is A $-23$

Given planes: $3x-2y+4z-7=0$ and $x+5y-2z+9=0$
Equation of plane containing the line of intersection is given by
$(3x-2y+4z-7)+\lambda (x+5y-2z+9)=0$
This plane passes through the point $(1,4,-3),$ so
$(3-8-12-7)+\lambda (1+20+6+9)=0\Rightarrow -24+36\lambda =0\Rightarrow \lambda =\frac{2}{3}$
Now, the required equation of plane is
$(3x-2y+4z-7)+\frac{2}{3}(x+5y-2z+9)=0\Rightarrow \frac{11x+4y+8z-3}{3}=0\Rightarrow -11x-4y-8z+3=0$
On comparing with $\alpha x+\beta y+\gamma z+3=0,$ we get
$\alpha =-11,\beta =-4,\gamma =-8\therefore \alpha +\beta +\gamma =-23$