If x = a,y = b,z = c is a solution of the sys | vedclass

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(C) The given system of linear equations is:$x + 8y + 7z = 0$  $(1)$$9x + 2y + 3z = 0$  $(2)$$x + y + z = 0$  $(3)$From equation $(3)$,$x = -y - z$. S

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$1$

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(C) The given system of linear equations is:$x + 8y + 7z = 0$  $(1)$$9x + 2y + 3z = 0$  $(2)$$x + y + z = 0$  $(3)$From equation $(3)$,$x = -y - z$. Substituting this into $(1)$ and $(2)$:$(-y - z) + 8y + 7z = 0 \implies 7y + 6z = 0 \implies z = -\frac{7}{6}y$$9(-y - z) + 2y + 3z = 0 \implies -7y - 6z = 0 \implies z = -\frac{7}{6}y$Since the equations are homogeneous and the determinant of the coefficient matrix is $0$,there are infinitely many solutions. Let $y = 6\lambda$. Then $z = -7\lambda$ and $x = -6\lambda - (-7\lambda) = \lambda$.So,$(a, b, c) = (\lambda, 6\lambda, -7\lambda)$.Since this point lies on the plane $x + 2y + z = 6$:$\lambda + 2(6\lambda) + (-7\lambda) = 6$$\lambda + 12\lambda - 7\lambda = 6$$6\lambda = 6 \implies \lambda = 1$.Thus,$(a, b, c) = (1, 6, -7)$.We need to find $2a + b + c = 2(1) + 6 + (-7) = 2 + 6 - 7 = 1$.

If $x = a$,$y = b$,$z = c$ is a solution of the system of linear equations $x + 8y + 7z = 0$,$9x + 2y + 3z = 0$,and $x + y + z = 0$ such that the point $(a, b, c)$ lies on the plane $x + 2y + z = 6$,then $2a + b + c$ equals

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