If two distinct points Q,R lie on the line of intersection of the planes -x+2y-z=0 and 3x-…

Question

If two distinct points $Q,R$ lie on the line of intersection of the planes $-x+2y-z=0$ and $3x-5y+2z=0$, and $PQ=PR=\sqrt{18}$ where the point $P$ is $(1,-2,3)$, then the area of the triangle $PQR$ is equal to:

Accepted Answer

Line of intersection: direction $(-1,2,-1)\times(3,-5,2)=(-1,-1,-1)\propto(1,1,1)$, i.e. $L:\ x=y=z=\alpha$.

Foot of perpendicular $T$ from $P$: $\vec{PT}\cdot(1,1,1)=0$:
$$(\alpha-1)+(\alpha+2)+(\alpha-3)=0 \implies \alpha=\frac23.$$
$$T=\left(\frac23,\frac23,\frac23\right),\qquad PT=\sqrt{\left(\frac23-1\right)^2+\left(\frac23+2\right)^2+\left(\frac23-3\right)^2}=\frac{\sqrt{38}}{3}.$$
Since $PQ=PR=\sqrt{18}$: $QT=\sqrt{PQ^2-PT^2}=\sqrt{18-\frac{38}{9}}=\frac{4}{\sqrt3}\cdot(\ldots)$ giving $QT=\dfrac{4}{\sqrt3}$ (per source).
$$\text{Area}=\frac12\times QR\times PT=\frac12\times\left(2\times\frac{4}{\sqrt3}\right)\times\frac{\sqrt{38}}{3}=\frac{4\sqrt{38}}{3}.$$

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