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

<p>Equation of plane containing the line $4ax - y + 5z - 7a = 0 = 2x - 5y - z - 3$ can be written as</p> <p>$4ax - y + 5z - 7a + \lambda (2x - 5y - z - 3) = 0$</p> <p>$(4a + 2\lambda )x - (1 + 5\lambda )y + (5 - \lambda )z - (7z + 3\lambda ) = 0$</p> <p>Which is coplanar with the line</p> <p>${{x - 4} \over 1} = {{y + 1} \over { - 2}} = {z \over 1}$</p> <p>$4(4a + 2\lambda ) + (1 + 5\lambda ) - (7a + 3\lambda ) = 0$</p> <p>$9a + 10\lambda + 1 = 0$ ..... (1)</p> <p>$(4a + 2\lambda )1 + (1 + 5\lambda )2 + 5 - \lambda = 0$</p> <p>$4a + 11\lambda + 7 = 0$ ...... (2)</p> <p>$a = 1,\,\lambda = - 1$</p> <p>Equation of plane is $x + 2y + 3z - 2 = 0$</p> <p>Intersection with the line</p> <p>${{x - 3} \over 7} = {{y - 2} \over { - 1}} = {{z - 3} \over { - 4}}$</p> <p>$(7t + 3) + 2( - t + 2) + 3( - 4t + 3) - 2 = 0$</p> <p>$- 7t + 14 = 0$</p> <p>$t = 2$</p> <p>So, the required point is $(17,0, - 5)$</p> <p>$\alpha + \beta + \gamma = 12$</p>