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Maths Question 14 – JEE-MAIN 2026

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Let a triangle PQR be such that P and Q lie on the line x+38=y−42=z+12 and are at a distance of 6 units from R (1,2,3). If (α,β,γ) is the centroid of △PQR, then α+β+γ is equal to :

Points P and Q lie on the given line and are equidistant from point R. This means P and Q are the intersection points of the line with a sphere centered at R with radius 6.

Step 1: Determine the coordinates of points P and Q✦ Active

Let the given line be L:x+38=y−42=z+12=λ. Any point on this line can be written as (8λ−3,2λ+4,2λ−1). Point R is (1,2,3). Since P and Q are at a distance of 6 units from R, we use the distance formula:

(8λ−3−1)2+(2λ+4−2)2+(2λ−1−3)2=62 (8λ−4)2+(2λ+2)2+(2λ−4)2=36 16(2λ−1)2+4(λ+1)2+4(λ−2)2=36 Dividing by 4: 4(4λ2−4λ+1)+(λ2+2λ+1)+(λ2−4λ+4)=9 16λ2−16λ+4+λ2+2λ+1+λ2−4λ+4=9 18λ2−18λ+9=9 18λ2−18λ=0⟹18λ(λ−1)=0 This yields λ=0 or λ=1. These values correspond to points P and Q. For λ=0, P=(−3,4,−1). For λ=1, Q=(5,6,1).
Step 2: Calculate the centroid coordinates○ Expand

The vertices of the triangle are P(−3,4,−1), Q(5,6,1), and R(1,2,3). The centroid G(α,β,γ) is given by:

α=−3+5+13=33=1 β=4+6+23=123=4 γ=−1+1+33=33=1 Thus, the centroid is G(1,4,1).
Step 3: Find the sum α+β+γ○ Expand

The required sum is α+β+γ:

α+β+γ=1+4+1=6
💡 Teacher's Secret Hint

Ensure all coordinates are correctly summed and divided by 3 for the centroid calculation.

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