JEE Main 2026 — Straight Lines Question with Solution

Let the coordinates of the points be $P(3\cos \alpha ,2\sin \alpha )$, $Q(x_Q,y_Q)$, and $R(x_R,y_R)$.

The centroid of $△PQR$ is given by:
$\left(\frac{3\cos \alpha +x_Q+x_R}{3},\frac{2\sin \alpha +y_Q+y_R}{3}\right)$

We are given that the centroid is $\left(2+\cos \alpha ,3+\frac{2}{3}\sin \alpha \right)$. Equating the coordinates, we get:
$\frac{3\cos \alpha +x_Q+x_R}{3}=2+\cos \alpha ⟹x_Q+x_R=6$
$\frac{2\sin \alpha +y_Q+y_R}{3}=3+\frac{2}{3}\sin \alpha ⟹y_Q+y_R=9$

Since the point $R$ lies on the line $x+y=5$, we have $x_R=5-y_R$.

Substituting $x_R$ into the equation for $x_Q$:
$x_Q=6-x_R=6-(5-y_R)=y_R+1$
And we already have $y_Q=9-y_R$.

The point $Q(x_Q,y_Q)$ lies on the circle $x^2+y^2-14x-14y+82=0$. Rewriting the circle's equation in standard form:
$(x-7{)}^2+(y-7{)}^2=16$

Substitute $x_Q$ and $y_Q$ into the circle's equation:
$(y_R+1-7{)}^2+(9-y_R-7{)}^2=16$
$(y_R-6{)}^2+(2-y_R{)}^2=16$

Expanding the terms:
y_R^2 - 12y_R + 36 + y_R^2 - 4y_R + 4 = 16
2y_R^2 - 16y_R + 40 = 16
2y_R^2 - 16y_R + 24 = 0
y_R^2 - 8y_R + 12 = 0

Factoring the quadratic equation:
$(y_R-2)(y_R-6)=0$

Thus, the possible ordinates for point $R$ are $y_R=2$ and $y_R=6$.

The sum of the ordinates of all possible points $R$ is $2+6=8$.

Answer: $8$