Let a triangle PQR be such that P and Q lie on the line x+38= y−42= z+12\; \dfrac{x+3}{8} = \; \dfrac{y-4}{2} = \; \dfrac{z+1}{2}8x+3=2y−4=2z+1 and are at a distance of 6 units from R(1,2,3)R(1,2,3)R(1,2,3). If (α, β, γ) is the centroid of ΔPQR\Delta PQRΔPQR, then α+β+γ\alpha + \beta + \gammaα+β+γ is equal to:
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