alleryPoints $A(3, 2, p), B(q, 8, -10), C(-2, -3, 1)$ are collinear. The values of $p$ and $q$ are:
A$p = 4,\ q = 9$
B$p = -4,\ q = 9$
C$p = 4,\ q = -9$
D$p = -4,\ q = -9$
Answer & Solution
Correct answer: B. $p = -4,\ q = 9$
Let $C$ divide $AB$ in ratio $t : 1$.
By section formula, equating components leads to $-3(t + 1) = 8t + 2$ ⇒ $t = -5/11$.
Substituting into the $x$- and $z$-coordinate equations yields $q = 9$ and $p = -4$.
(The negative $t$ confirms external division — $C$ lies outside segment $AB$.)
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