$\begin{array}{1 1} x=3 \\y=3 \\z=3 \\ x+y+x=3\end{array} $

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- Equation of a plane,whose intercepts are $a,b,c$ is given by $\large\frac{x}{a}$$+\large\frac{y}{b}$$+\large\frac{z}{c}$$=1$

Step 1:

We know that the equation of the plane $ZOX$ is $y=0$

Any plane which is parallel to it is of the form $y=a$

Where $a$ is the $y$-intercept of the plane.

Step 2:

But it is given $a=3$

Therefore the equation of the required plane is $y=3$

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