JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Three Dimensional Geometry

Let the plane be $ax+by+cz=d$. Through $(0,-1,0)$ gives $-b=d$ and through $(0,0,1)$ gives $c=d$, so the plane is $ax-dy+dz=d$; dividing by $d$ (nonzero, else the angle could not be $45^{\circ }$) it is $px-y+z=1$ with $p=a/d$. Its normal $(p,-1,1)$ makes $45^{\circ }$ with the normal $(0,1,-1)$ of $y-z+5=0$: $\cos 45^{\circ }=\frac{|-1-1|}{\sqrt{p^2+2}\sqrt{2}}=\frac{2}{\sqrt{2}\sqrt{p^2+2}}$. Setting this to $\frac{1}{\sqrt{2}}$ gives $p^2+2=4$, so $p=\pm \sqrt{2}$. Taking $p=-\sqrt{2}$, the plane $-\sqrt{2}x-y+z=1$ is satisfied by $(\sqrt{2},1,4)$ since $-\sqrt{2}\cdot \sqrt{2}-1+4=-2-1+4=1$.