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Maths Question 15 – JEE-MAIN 2026

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If the distance of the point (a,2,5) from the image of the point (1,2,7) in the line x1=y−11=z−22 is 4, then the sum of all possible values of a is equal to :

To find the image of a point in a line, first determine the foot of the perpendicular from the given point to the line.

● Video Walkthrough
Step 1: Find the Foot of the Perpendicular✦ Active

Let the given line be L:x1=y−11=z−22=λ. A general point on the line is M(λ,λ+1,2λ+2). The given point is Q(1,2,7). The direction vector of the line is d→=(1,1,2). The vector QM→ is perpendicular to d→. Thus, QM→⋅d→=0.

QM→=(λ−1,(λ+1)−2,(2λ+2)−7)=(λ−1,λ−1,2λ−5) (λ−1)(1)+(λ−1)(1)+(2λ−5)(2)=0 2(λ−1)+4λ−10=0 2λ−2+4λ−10=0 6λ−12=0⟹λ=2

Substitute λ=2 into the coordinates of M: M(2,2+1,2(2)+2)=(2,3,6). This is the foot of the perpendicular.

Step 2: Determine the Image Point○ Expand

Let Q′(x′,y′,z′) be the image of Q(1,2,7) in the line L. The foot of the perpendicular M(2,3,6) is the midpoint of Q and Q′. Using the midpoint formula:

1+x′2=2⟹1+x′=4⟹x′=3 2+y′2=3⟹2+y′=6⟹y′=4 7+z′2=6⟹7+z′=12⟹z′=5

So, the image point Q′ is (3,4,5).

Step 3: Apply Distance Condition and Sum 'a' Values○ Expand

The distance between point P(a,2,5) and the image point Q′(3,4,5) is given as 4. Using the distance formula:

D=(a−3)2+(2−4)2+(5−5)2 4=(a−3)2+(−2)2+02 4=(a−3)2+4

Square both sides to solve for a:

16=(a−3)2+4 (a−3)2=12 a−3=±12 a−3=±23 a=3±23

The possible values of a are 3+23 and 3−23. The sum of all possible values of a is:

(3+23)+(3−23)=3+3=6
💡 Teacher's Secret Hint

Remember to consider both positive and negative roots when solving for (a−3). The sum of roots for a quadratic equation x2−Sx+P=0 is S. Here, (a−3)2=12 is equivalent to a2−6a+9=12, or a2−6a−3=0. The sum of roots is −(−6)/1=6.

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