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 $4 a x-y+5 z-7 a=0$ $=2 x -5 y - z -3, \quad a \in R$ at the point $P(\alpha, \beta, \gamma)$. Then the value of $\alpha+\beta+\gamma$ equals ______

Answer 1

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