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The locus of the point $P(\mathbf{r})$ which encloses a triangle $A B P$ of area l sq. unit with the fixed points $A(\hat{\mathbf{i}})$ and $B(\hat{\mathbf{j}})$ is
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$(x+y-1)^2+2 z^2=4$
Let point $P(\mathbf{r})$ is $x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$ and $A(\hat{\mathbf{i}})$ and $B(\hat{\mathbf{j}})$
$\begin{aligned} & \qquad \quad \mathbf{A P}=(x-1) \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}} \Rightarrow \mathbf{A B}=-\hat{\mathbf{i}}+\hat{\mathbf{j}} \\ & \text { Area of } \triangle A B P=\frac{1}{2}|\mathbf{A P} \times \mathbf{A B}|\end{aligned}$
$\begin{aligned} & \mathbf{A P} \times \mathbf{A B}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ x-1 & y & z \\ -1 & 1 & 0\end{array}\right|=-z \hat{\mathbf{i}}+z \hat{\mathbf{j}}+(x+y-1) \hat{\mathbf{k}} \\ & |\mathbf{A} \mathbf{P} \times \mathbf{A B}|=\sqrt{2 z^2+(x+y-1)^2} \\ & \therefore \quad 1=\frac{1}{2} \sqrt{2 z^2+(x+y-1)^2} \quad[\because \text { Area of } \triangle A B P=1] \\ & =(x+y-1)^2+2 z^2=4 \\ & \end{aligned}$
$\begin{aligned} & \qquad \quad \mathbf{A P}=(x-1) \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}} \Rightarrow \mathbf{A B}=-\hat{\mathbf{i}}+\hat{\mathbf{j}} \\ & \text { Area of } \triangle A B P=\frac{1}{2}|\mathbf{A P} \times \mathbf{A B}|\end{aligned}$
$\begin{aligned} & \mathbf{A P} \times \mathbf{A B}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ x-1 & y & z \\ -1 & 1 & 0\end{array}\right|=-z \hat{\mathbf{i}}+z \hat{\mathbf{j}}+(x+y-1) \hat{\mathbf{k}} \\ & |\mathbf{A} \mathbf{P} \times \mathbf{A B}|=\sqrt{2 z^2+(x+y-1)^2} \\ & \therefore \quad 1=\frac{1}{2} \sqrt{2 z^2+(x+y-1)^2} \quad[\because \text { Area of } \triangle A B P=1] \\ & =(x+y-1)^2+2 z^2=4 \\ & \end{aligned}$
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