JEE Main 2019 — Mathematics Algebra

The system of equations ${a}_{1}x+{b}_{1}y+{c}_{1}z=0,{a}_{2}x+{b}_{2}y+{c}_{2}z=0$a1​x+b1​y+c1​z=0,a2​x+b2​y+c2​z=0 and ${a}_{3}x+{b}_{3}y+{c}_{3}z=0$a3​x+b3​y+c3​z=0 has a non-trivial solution, if

$$|\begin{matrix}{a}_{1} & {b}_{1} & {c}_{1} \\ {a}_{2} & {b}_{2} & {c}_{2} \\ {a}_{3} & {b}_{3} & {c}_{3}\end{matrix}|=0.$$∣a1​a2​a3​​b1​b2​b3​​c1​c2​c3​​∣=0.

Hence, for the given lines $2x+3y-z=0,x+ky-2z=0$2x+3y−z=0,x+ky−2z=0 and $2x-y+z=0,$2x−y+z=0, we have

$$|\begin{matrix}2 & 3 & -1 \\ 1 & k & -2 \\ 2 & -1 & 1\end{matrix}|=0$$∣212​3k−1​−1−21​∣=0

$$\Rightarrow 2(k-2)-3(1+4)-1(-1-2k)=0$$⇒2(k−2)−3(1+4)−1(−1−2k)=0

$$\Rightarrow 2k-4-15+1+2k=0$$⇒2k−4−15+1+2k=0

$$\Rightarrow k=\frac{9}{2}$$⇒k=29​

Thus, the lines are

$$2x+3y-z=0...(1)$$2x+3y−z=0...(1)

$$x+\frac{9}{2}y-2z=0...(2)$$x+29​y−2z=0...(2)

$$2x-y+z=0...(3)$$2x−y+z=0...(3)

$$(1)-(3)\Rightarrow 4y-2z=0$$(1)−(3)⇒4y−2z=0

$$2y=z...(4)$$2y=z...(4)

$$\frac{y}{z}=\frac{1}{2}...(5)$$zy​=21​...(5)

put $z$z from eq $(4)$(4) into $(1)$(1)

$$2x+3y-2y=0$$2x+3y−2y=0

$$2x+y=0$$2x+y=0

$$\frac{x}{y}=-\frac{1}{2}...(6)$$yx​=−21​...(6)

From eq $(4)$(4) and $(6)$(6)

$$\frac{z}{x}=-4$$xz​=−4

Hence, $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+k=-\frac{1}{2}+\frac{1}{2}-4+\frac{9}{2}=\frac{1}{2}.$yx​+zy​+xz​+k=−21​+21​−4+29​=21​.