JEE Main 2026 · 6 Apr · Shift 1 · Mathematics Q15

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MathematicsSingle correct+4 / -1Question 15 of 75

Let the image of the point $P(1,6,a)$ in the line $L:\dfrac x1=\dfrac{y-1}2=\dfrac{z-a+1}b$ , $b>0$ , be $\left(\dfrac a3,0,a+c\right)$ . If $S(\alpha,\beta,\gamma)$ , $\alpha>0$ , is the point on L such that the distance of S from the foot of perpendicular from the point P on L is $2\sqrt{14}$ , then $\alpha+\beta+\gamma$ is equal to:

Answer and worked solution

Official answer: C

Midpoint on L. The midpoint of P and its image is $\left(\tfrac{1+a/3}2,\,3,\,\tfrac{2a+c}2\right)$ . From $\tfrac x1=\tfrac{y-1}2=1$ : $a=3$ . Then the z-coordinate gives $b=\tfrac{c+2}2$ .

Perpendicularity. $(0,6,-c)\cdot(1,2,b)=0$ , so $bc=12$ . This gives $c^2+2c-24=0$ , so $c=4$ and $b=3$ .

Foot of the perpendicular. The midpoint, $(1,3,5)$ .

Point S. Along $(1,2,3)$ (length $\sqrt{14}$ ), $S=(1,3,5)\pm2(1,2,3)$ . With $\alpha>0$ , $S=(3,7,11)$ , and the sum is 21.

(C)

Topics:Equation of a line in space Angles and distances

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JEE Main 2026 · 6 Apr · Shift 1, Mathematics question 15