UPSC CDS 2 2021 Math Paper

Concept:The equality of the three ratios is used to set each equal to a common constant $k$, leading to the conclusion $k=0$, which gives proportional relationships between $x,y,z$ and $a,b,c$.Explanation:Let $\frac{ay-bx}{c} = \frac{cx-az}{b} = \frac{bz-cy}{a} = k$.Multiply numerator and denominator of each fraction by $c$, $b$, $a$ respectively:$\frac{acy-bcx}{c^{2}} = \frac{bcx-abz}{b^{2}} = \frac{abz-acy}{a^{2}} = k$.This gives three equations:$acy - bcx = c^{2}k$$bcx - abz = b^{2}k$$abz - acy = a^{2}k$Add them: $(acy - bcx) + (bcx - abz) + (abz - acy) = (a^{2}+b^{2}+c^{2})k$The left side simplifies to $0$, so $0 = (a^{2}+b^{2}+c^{2})k$.Since $a^{2}+b^{2}+c^{2} > 0$ (unless all zero, but then the original fractions are undefined), we get $k = 0$.Thus $acy = bcx$, $bcx = abz$, $abz = acy$.From $acy = bcx$, cancel $c$ (nonzero) to get $ay = bx$, i.e. $\frac{x}{a} = \frac{y}{b}$.From $bcx = abz$, cancel $b$ to get $cx = az$, i.e. $\frac{x}{a} = \frac{z}{c}$.So $\frac{x}{a} = \frac{y}{b} = \frac{z}{c} = r$ (say), implying $x = ar$, $y = br$, $z = cr$.Check condition 1: $\frac{x}{a} = \frac{y}{b}$ is true from above.Check condition 2: $\frac{x+y+z}{a+b+c} = \frac{ar+br+cr}{a+b+c} = r$, and $\frac{z}{c} = \frac{cr}{c} = r$, so condition 2 is also true.Hence both statements are correct.Answer:Both statements 1 and 2 are correct. The correct code is C.