JEE Main 2026 — Three Dimensional Geometry Question with Solution

Let $P(1,2,7)$ be the given point and the given line be $L:\frac{x}{1}=\frac{y-1}{1}=\frac{z-2}{2}=\lambda$.

Any point on the line can be written as $F(\lambda ,\lambda +1,2\lambda +2)$.

If $F$ is the foot of the perpendicular from $P$ to the line, the direction ratios of $PF$ are $(\lambda -1,\lambda -1,2\lambda -5)$.

Since $PF$ is perpendicular to the line, the dot product of their direction ratios is zero:

$1(\lambda -1)+1(\lambda -1)+2(2\lambda -5)=0$

$6\lambda -12=0\Rightarrow \lambda =2$

Thus, the foot of the perpendicular is $F(2,3,6)$.

Let the image of $P$ in the line be $P^{\prime }(x^{\prime },y^{\prime },z^{\prime })$. Since $F$ is the midpoint of $P$ and $P^{\prime }$:

$\frac{x^{\prime }+1}{2}=2\Rightarrow x^{\prime }=3$

$\frac{y^{\prime }+2}{2}=3\Rightarrow y^{\prime }=4$

$\frac{z^{\prime }+7}{2}=6\Rightarrow z^{\prime }=5$

The image point is $P^{\prime }(3,4,5)$.

The distance of the point $Q(a,2,5)$ from $P^{\prime }(3,4,5)$ is given as $4$.

$\sqrt{(a-3{)}^2+(2-4{)}^2+(5-5{)}^2}=4$

$(a-3{)}^2+4=16$

$(a-3{)}^2=12$

$a^2-6a-3=0$

The sum of all possible values of $a$ is the sum of the roots of this quadratic equation, which is $6$.

Answer: $6$