$2(x-1) + 3(y-2) - (z-3) = 0$
$$2x + 3y - z = 2 + 6 - 3 = 5$$
Verified 30 May 2026.
The equation of the plane passing through $(1, 2, 3)$ and perpendicular to the vector $2\hat{i} + 3\hat{j} - \hat{k}$ is:
$2x + 3y - z = 5$
$2x + 3y - z = 1$
$2x - 3y + z = 5$
$x + 2y - 3z = 5$
$2(x-1) + 3(y-2) - (z-3) = 0$
$$2x + 3y - z = 2 + 6 - 3 = 5$$
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