If the system of linear equations, x + y + z = 6 x + 2y + 3z = 10...

$x + y + z = 6 . . . \left(1\right)<br />$x + 2y + 3z = 10 . . . \left(2\right)$<br />$3x + 2y + \lambdaz = \mu \ldots \left(3\right)

From \left(1\right)\text{and }\left(2\right),<br />Ifz = 0 ⇒ x + y = 6 and x + 2y = 10$,$⇒ y = 4, x = 2$<br />If$y = 0 ⇒ x + z = 6 and x + 3z = 10$,$⇒ z = 2, x = 4$</p><p>So,$3x + 2y + \lambdaz = \mu$, must pass through$( 2,4, 0 ) and ( 4,0, 2 )$.</p><p>$6 + 8 = \mu ⇒ \mu = 14$<br />And$12 + 2\lambda = \mu ⇒ \lambda = 1$</p><p>So,$\mu - {\lambda}^{2} = 14 - 1= 13$</p>$