CUET UG 2025 — Mathematics Geometry

Given points:

• A = (-1, -1, 2)
• B = (2, m, 5)
• C = (3, 11, 6)

For three points to be collinear, vectors AB and AC must be parallel.

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Vector AB = B - A

$$AB = (2 - (-1), m - (-1), 5 - 2)$$AB=(2−(−1),m−(−1),5−2)

$$AB = (3, m+1, 3)$$AB=(3,m+1,3)

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Vector AC = C - A

$$AC = (3 - (-1), 11 - (-1), 6 - 2)$$AC=(3−(−1),11−(−1),6−2)

$$AC = (4, 12, 4)$$AC=(4,12,4)

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For collinearity, $AB = k \times AC$AB=k×AC for some constant $k$k.

$$(3, m+1, 3) = k(4, 12, 4)$$(3,m+1,3)=k(4,12,4)

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From the first component:

$$3 = 4k$$3=4k

$$k = \dfrac{3}{4}$$k=43​

From the third component:

$$3 = 4k$$3=4k

$$k = \dfrac{3}{4}$$k=43​

Both components give the same value of $k$k.

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Using the second component:

$$m + 1 = 12k$$m+1=12k

$$m + 1 = 12 \times \dfrac{3}{4}$$m+1=12×43​

$$m + 1 = 9$$m+1=9

$$m = 8$$m=8

Therefore, $m = 8$m=8.