(A) Given that $x, y, z \in \mathbb{R}$ and $x^4+y^4+z^4=0$. Since the sum of even powers of real numbers is zero,each term must be zero individually:

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(A) Given that $x, y, z \in \mathbb{R}$ and $x^4+y^4+z^4=0$. Since the sum of even powers of real numbers is zero,each term must be zero individually: $x^4=0, y^4=0, z^4=0$. This implies $x=0, y=0, z=0$. Now,substitute $x=0, y=0, z=0$ into the determinant: $\Delta = \left|\begin{array}{ccc}1 & (0)(0) & (0)(0) \\ (0)(0) & 1 & (0)(0) \\ (0)(0) & (0)(0) & 1\end{array}\right| = \left|\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|$. The determinant of the identity matrix is $1(1-0) - 0 + 0 = 1$. Thus,the value is $1$.

Class 12 Mathematics 3 and 4 .Determinants and Matrices Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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If $x^4+y^4+z^4=0$ then,$\left|\begin{array}{ccc}1 & xy & yz \\ zx & 1 & xy \\ yz & zx & 1\end{array}\right|=$ . . . . . . . $(\because x, y, z \in \mathbb{R})$

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A
$1$
B
$x+y+z+3$
C
$xyz+2$
D
$0$

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3 and 4 .Determinants and Matrices

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Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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If $x^4+y^4+z^4=0$ then,$\left|\begin{array}{ccc}1 & xy & yz \\ zx & 1 & xy \\ yz & zx & 1\end{array}\right|=$ . . . . . . . $(\because x, y, z \in \mathbb{R})$

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