JEE Main 2026 — Mathematics Vectors & 3D Geometry

The equation of the given line is $\dfrac{x+3}{8} = \dfrac{y-4}{2} = \dfrac{z+1}{2}$8x+3​=2y−4​=2z+1​.

Let $\dfrac{x+3}{4} = \dfrac{y-4}{1} = \dfrac{z+1}{1} = \lambda$4x+3​=1y−4​=1z+1​=λ.

Any point on this line can be taken as $S(4\lambda - 3, \lambda + 4, \lambda - 1)$S(4λ−3,λ+4,λ−1).

Since the points $P$P and $Q$Q lie on this line and are at a distance of $6$6 units from $R(1, 2, 3)$R(1,2,3), we have $RS^2 = 36$RS2=36.

$$(4\lambda - 3 - 1)^2 + (\lambda + 4 - 2)^2 + (\lambda - 1 - 3)^2 = 36$$(4λ−3−1)2+(λ+4−2)2+(λ−1−3)2=36

$$(4\lambda - 4)^2 + (\lambda + 2)^2 + (\lambda - 4)^2 = 36$$(4λ−4)2+(λ+2)2+(λ−4)2=36

$$16(\lambda^2 - 2\lambda + 1) + (\lambda^2 + 4\lambda + 4) + (\lambda^2 - 8\lambda + 16) = 36$$16(λ2−2λ+1)+(λ2+4λ+4)+(λ2−8λ+16)=36

$$18\lambda^2 - 36\lambda + 36 = 36$$18λ2−36λ+36=36

$$18\lambda(\lambda - 2) = 0 \Rightarrow \lambda = 0 \text{ or } \lambda = 2$$18λ(λ−2)=0⇒λ=0 or λ=2

For $\lambda = 0$λ=0, the point is $P(-3, 4, -1)$P(−3,4,−1).

For $\lambda = 2$λ=2, the point is $Q(5, 6, 1)$Q(5,6,1).

The centroid $(\alpha, \beta, \gamma)$(α,β,γ) of $\triangle PQR$△PQR is given by:

$$\alpha = \dfrac{-3 + 5 + 1}{3} = 1$$α=3−3+5+1​=1

$$\beta = \dfrac{4 + 6 + 2}{3} = 4$$β=34+6+2​=4

$$\gamma = \dfrac{-1 + 1 + 3}{3} = 1$$γ=3−1+1+3​=1

Therefore, $\alpha + \beta + \gamma = 1 + 4 + 1 = 6$α+β+γ=1+4+1=6.

Answer: $6$6