Match List-I with List-II List-I (points)List-II (the points form) (A…

The correct answer is Option (1) → (A)-(IV), (B)-(I), (C)-(III), (D)-(II)

For points P, Q, R, find the side vectors \(\overrightarrow{PQ}\), \(\overrightarrow{PR}\), \(\overrightarrow{QR}\). The points are collinear if two side vectors are parallel; a zero scalar product of two sides gives a right angle; the lengths decide isosceles or equilateral.

(A) (1, 2, 3), (2, 4, 5), (4, 8, 9)

This matches (IV) no triangle.

    \(\overrightarrow{PQ} = \hat{\imath}+2\hat{\jmath}+2\hat{k}, \quad \overrightarrow{PR} = 3\hat{\imath}+6\hat{\jmath}+6\hat{k} = 3\,\overrightarrow{PQ}\)

The points lie on one line.

(B) (1, 1, 1), (3, 2, 1), (0, 3, 2)

This matches (I) a right-angled triangle.

    \(\overrightarrow{PQ} = 2\hat{\imath}+\hat{\jmath}, \quad \overrightarrow{PR} = -\hat{\imath}+2\hat{\jmath}+\hat{k}\)

    \(\overrightarrow{PQ}\cdot\overrightarrow{PR} = -2+2+0 = 0\)

The angle at P is a right angle (the sides are \(\sqrt{5}\), \(\sqrt{6}\), \(\sqrt{11}\)).

(C) (2, 1, −1), (4, 2, 1), (3, 3, 1)

This matches (III) an isosceles triangle that is not right-angled.

    \(|\overrightarrow{PQ}| = |2\hat{\imath}+\hat{\jmath}+2\hat{k}| = 3, \quad |\overrightarrow{PR}| = |\hat{\imath}+2\hat{\jmath}+2\hat{k}| = 3\)

    \(|\overrightarrow{QR}| = |-\hat{\imath}+\hat{\jmath}| = \sqrt{2}\)

Two sides are equal, and \(3^2+(\sqrt{2})^2 \ne 3^2\), \(3^2+3^2 \ne 2\), so no angle is a right angle.

(D) (3, 1, 1), (1, 3, 1), (1, 1, 3)

This matches (II) an equilateral triangle.

    \(|\overrightarrow{PQ}| = |-2\hat{\imath}+2\hat{\jmath}| = 2\sqrt{2}, \quad |\overrightarrow{PR}| = |-2\hat{\imath}+2\hat{k}| = 2\sqrt{2}\)

    \(|\overrightarrow{QR}| = |-2\hat{\jmath}+2\hat{k}| = 2\sqrt{2}\)