Question

Consider the system of equations
$x+y+z=4$
$2x+y+3z=6$
$x+2y+pz=q$
If the system of equations has no solution for $p=p^{\prime }$ and has infinite solutions for $q=q^{\prime },$ then the value of $(p^{\prime }{)}^2+(q^{\prime }{)}^2$ is

Answer 1

$x+y+z=4...\left(1\right)$
$2x+y+3z=6...\left(2\right)$
$x+2y+pz=q...\left(3\right)$

Solving $\left(1\right)$ and $\left(2\right),$
$x=2-2z$ and $y=2+z$
Putting in equation $\left(3\right),$ we get
$pz=q-6$

For unique solution, $p\ne 0,q\in R$
For no solution, we must have $p=0,q\ne 6$
For infinite solution, $p=0$ and $q=6$

$\therefore p^{\prime }=0$ and $q^{\prime }=6$
$\Rightarrow (p^{\prime }{)}^2+(q^{\prime }{)}^2=0+36=36$