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Matrices & Determinants: Pqr System Equations Non Trivial Solution Equals

JEE Maths question with a full step-by-step solution.

Question
If pqr≠0pqr \neq 0 and the system of equations (p+a)x+by+cz=0(p+a)x+by+cz=0, ax+(q+b)y+cz=0ax+(q+b)y+cz=0, ax+by+(r+c)z=0ax+by+(r+c)z=0 has a non-trivial solution, then ap+bq+cr\dfrac{a}{p}+\dfrac{b}{q}+\dfrac{c}{r} equals:
A11
B22
C00
D−1-1correct
Solution
Step 1: Write the condition for a non-trivial solution
∣p+abcaq+bcabr+c∣=0\begin{vmatrix} p+a & b & c \\ a & q+b & c \\ a & b & r+c \end{vmatrix} = 0
 Step 2: Apply R1→R1−R2R_1 \to R_1-R_2 and R2→R2−R3R_2 \to R_2-R_3, then factor After subtracting consecutive rows:
∣p−q00q−rabr+c∣=0\begin{vmatrix} p & -q & 0 \\ 0 & q & -r \\ a & b & r+c \end{vmatrix} = 0
 Expanding: p[q(r+c)+rb]+q[0⋅(r+c)+ra]=0p\bigl[q(r+c)+rb\bigr]+q\bigl[0\cdot(r+c)+ra\bigr] = 0
pq(r+c)+prb+qra=0pq(r+c)+prb+qra = 0
 Step 3: Divide by pqrpqr
r+cr+bq+ap=0  ⟹  1+cr+bq+ap=0\frac{r+c}{r} + \frac{b}{q} + \frac{a}{p} = 0 \implies 1 + \frac{c}{r} + \frac{b}{q} + \frac{a}{p} = 0
ap+bq+cr=−1\frac{a}{p}+\frac{b}{q}+\frac{c}{r} = -1
 Answer: (4)
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