The locus of the equation {x^2} + {y^2} + {z^ | vedclass

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Question

(A) Given the equation is ${x^2} + {y^2} + {z^2} + 1 = 0$. For any real values of $x, y, z$,the squares ${x^2}, {y^2}, {z^2}$ are always non-negative,

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An empty set

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(A) Given the equation is ${x^2} + {y^2} + {z^2} + 1 = 0$. For any real values of $x, y, z$,the squares ${x^2}, {y^2}, {z^2}$ are always non-negative,i.e.,${x^2} \ge 0, {y^2} \ge 0, {z^2} \ge 0$. Therefore,the sum ${x^2} + {y^2} + {z^2} \ge 0$. Adding $1$ to both sides,we get ${x^2} + {y^2} + {z^2} + 1 \ge 1$. Since the expression is always greater than or equal to $1$,it can never be equal to $0$. Thus,there are no real coordinates $(x, y, z)$ that satisfy this equation. Hence,the locus is an empty set.

The locus of the equation ${x^2} + {y^2} + {z^2} + 1 = 0$ is

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