Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The midpoint of the segment joining $M(-4,2,1)$ and $N(2,-4,3)$ is given by $\left(\frac{-4+2}{2},\frac{2-4}{2},\frac{1+3}{2}\right)=(-1,-1,2)$.

The direction ratios of the segment $MN$ are $(2-(-4),-4-2,3-1)=(6,-6,2)$.

The normal vector to the plane $P$ is proportional to the direction ratios of $MN$, which can be simplified to $(3,-3,1)$.

The equation of the plane $P$ passing through $(-1,-1,2)$ with normal vector $(3,-3,1)$ is:

$3(x+1)-3(y+1)+1(z-2)=0$

$3x-3y+z-2=0$

The equation of the line $L$ is $\frac{x-1}{1}=\frac{y+2}{3}=\frac{z-7}{-1}=t$.

Any point on the line $L$ can be written as $(t+1,3t-2,-t+7)$.

Substituting this point into the equation of the plane $P$ gives:

$3(t+1)-3(3t-2)+(-t+7)-2=0$

$3t+3-9t+6-t+7-2=0$

$-7t+14=0\Rightarrow t=2$

Substituting $t=2$ into the coordinates of the point on the line gives the intersection point $S(\alpha ,\beta ,\gamma )$:

$\alpha =2+1=3$

$\beta =3(2)-2=4$

$\gamma =-2+7=5$

Therefore, $\alpha +\beta +\gamma =3+4+5=12$.

Answer: $12$