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

Equation of plane

$4\mathrm{ax}-\mathrm{y}+5\mathrm{z}-7\mathrm{a}+\mathrm{\lambda }(2\mathrm{x}-5\mathrm{y}-\mathrm{z}-3)=0$

this satisfy (4, -1, 0)

$\begin{array}{l}16\mathrm{a}+1-7\mathrm{a}+\mathrm{\lambda }(8+5-3)=0\\ 9\mathrm{a}+1+10\mathrm{\lambda }=0 ....\left(1\right)\end{array}$

Normal vector of the plane A is $(4\mathrm{a}+2\mathrm{\lambda }, -1-5\mathrm{\lambda }, 5-\mathrm{\lambda })$ vector along the line which contained the plane A is

$\begin{array}{l}\mathrm{i}-2\mathrm{j}+\mathrm{k}\\ \therefore 4\mathrm{a}+2\mathrm{\lambda }+2+10\mathrm{\lambda }+5-\mathrm{\lambda }=0\\ 11\mathrm{\lambda }+4\mathrm{a}+7=0 \dots \dots \left(2\right)\end{array}$

Solve (1) and (2) to get a = 1, $\mathrm{\lambda }$ = -1

Now equation of plane

$\mathrm{x}+2\mathrm{y}+3\mathrm{z}-2=0$

Let the point in the line $\frac{\mathrm{x}-3}{7}=\frac{\mathrm{y}-2}{-1}=\frac{\mathrm{z}-3}{-4}=\mathrm{t}$ is $(7\mathrm{t}+3,-\mathrm{t}+2,-4\mathrm{t}+3)$ satisfy the equation of plane A

$$\begin{gathered} 7\mathrm{t}+3-2\mathrm{t}+4+9-12\mathrm{t}-2=0 \\ \mathrm{t}=2 \\ \mathrm{So} \mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }=2\mathrm{t}+8=12 \end{gathered}$$