Question

Let the line $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z-3}{-4}$ intersect the plane containing the lines $\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$ and $4ax-y+5z-7a=0$ $=2x-5y-z-3,a\in R$ at the point $P(\alpha ,\beta ,\gamma )$. Then the value of $\alpha +\beta +\gamma$ equals ______

Answer 1

Answer: 12

Equation of plane
$4ax-y+5z-7a+\lambda (2x-5y-z-3)=0$
this satisfy $(4,-1,0)$
$16a+1-7a+\lambda (8+5-3)=0$
$9a+1+10\lambda =\mathrm{0......}$(1)
Normal vector of the plane $A$ is $(4a+2\lambda ,-1-5\lambda ,5-\lambda )$ vector along the line which contained the plane $A$ is
$i-2j+k$
$\therefore 4a+2\lambda +2+10\lambda +5-\lambda =0$
$11\lambda +4a+7=0\dots .(2)$
Solve (1) and (2) to get $a=1,\lambda =-1$
Now equation of plane
$x+2y+3z-2=0$
Let the point in the line $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z-3}{-4}=t$
is $(7t+3,-t+2,-4t+3)$ satisfy the equation of plane $A$
$7t+3-2t+4+9-12t-2=0$
$t=2$
So $\alpha +\beta +\gamma =2t+8=12$