Finding orthocentre of $\Delta$ formed by intercepts of plane $3x+y+3z=9$

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Question:
The plane $3x+y+3z=9$ intersects the co-ordinate axes at $A, B, C$. Find orthocentre $(H)$ of $\Delta ABC$.

We see $\Delta ABC$ is isoceles, so let's assume $H = (h, k, h)$, as vertices $A$ and $C$ are symmetrical about $y$-axis.

It follows that, $AH \perp BC$, then $$(h-3, k, h)\cdot(0, -9,3)=0$$ $$\Rightarrow h=3k$$

Also, as $H$ lies on plane itself, then $$3h+k+3h=9$$ $$\Rightarrow 6h=9-k$$

On solving, we get, $$H=(27/19, 9/19,27/19)$$

Is this a proper way of finding orthocentre? (Bcoz, my answer isn't matching with the key)


The correct answer is:

(1,1,3)