Let $\mathrm{P}_{1}$ be the plane $3 x-y-7 z=11$ and $\mathrm{P}_{2}$ be the plane passing through the points $(2,-1,0),(2,0,-1)$, and $(5,1,1)$. If the foot of the perpendicular drawn from the point $(7,4,-1)$ on the line of intersection of the planes $P_{1}$ and $P_{2}$ is $(\alpha, \beta, \gamma)$, then $\alpha+\beta+\gamma$ is equal to ___________.

Equation of plane $\mathrm{P}_2$ passing through $(2,-1,0),(2,0$, $-1)$ and $(5,1,1)$ is <br/><br/>$$ \begin{aligned} & \left|\begin{array}{ccc} x-5 & y-1 & z-1 \\ 3 & 2 & 1 \\ 3 & 1 & 2 \end{array}\right|=0 \\\\ & \Rightarrow(x-5)(4-1)-(y-1)(6-3)+(z-1)(3-6)=0 \\\\ & \Rightarrow 3 x-15-3 y+3-3 z+3=0 \\\\ & \Rightarrow 3 x-3 y-3 z-9=0 \\\\ & \Rightarrow x-y-z=3 .......(i) \end{aligned} $$ <br/><br/>Now, direction ratios of line of intersection of $P_1$ and $\mathrm{P}_2$ is <br/><br/>$$ \begin{aligned} & \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & -1 \\ 3 & -1 & -7 \end{array}\right| \\\\ & =\hat{i}(7-1)-\hat{j}(-7+3)+\hat{k}(-1+3) \\\\ & =6 \hat{i}+4 \hat{j}+2 \hat{k} \end{aligned} $$ <br/><br/>At $z=0, x-y=3$ [from (i)] <br/><br/>$3 x-y=11$ <br/><br/>On solving, we get <br/><br/>$x=4 \text { and } y=1$ <br/><br/>So, equation of line is <br/><br/>$\frac{x-4}{6}=\frac{y-1}{4}=\frac{z-2}{6}=k$ <br/><br/>$$ \begin{aligned} & \therefore(\alpha, \beta, \gamma)=(6 k+4,4 k+1,2 k) \\\\ & \Rightarrow(6)(\alpha-7)+4(\beta-4)+2(\gamma+1)=0 \\\\ & \Rightarrow 6(6 k+4-7)+4(4 k+1-4)+2(2 k+1)=0 \\\\ & \Rightarrow 36 k-18+16 k-12+4 k+4=0 \\\\ & \Rightarrow 56 k=26 \Rightarrow k=\frac{1}{2} \\\\ & \text { So, } \alpha=7, \beta=3 \text { and } \gamma=1 \\\\ & \therefore \alpha+\beta+\gamma=7+3+1=11 \end{aligned} $$