Q.Find the vector equation of the plane which makes intercepts 1, 1, 1 on the co-ordinates axes.

[!TLDR] The three intercepts $(1,0,0)$, $(0,1,0)$, $(0,0,1)$ are three non-collinear points of the plane; either apply Theorem 6.11 to them or recognise the direct intercep …

A plane making intercepts $1,1,1$ on the axes passes through the three points $A(1,0,0)$, $B(0,1,0)$, $C(0,0,1)$ — three non-collinear points, so Theorem 6.11 applies.

Here $\vec a=\hat i$, $\vec b=\hat j$, $\vec c=\hat k$, so $\vec b-\vec a = -\hat i+\hat j$ and $\vec c-\vec a=-\hat i+\hat k$.

$\hat i$-component: $(1)(1)-(0)(0)=1$

$\hat j$-component: $-[(-1)(1)-(0)(-1)] = -(-1-0)=1$

$\hat k$-component: $(-1)(0)-(1)(-1) = 0+1=1$

$(\vec b-\vec a)\times(\vec c-\vec a) = \hat i+\hat j+\hat k$ …

1. Convert the given intercepts $1,1,1$ into the three axis points $(1,0,0),(0,1,0),(0,0,1)$.
2. Apply Theorem 6.11 (plane through three non-collinear points): compute $(\vec b-\vec a)\times(\vec c-\vec a)$ as the normal. …

Misreading 'intercepts 1,1,1' as the single point $(1,1,1)$ rather than three separate axis points $(1,0,0),(0,1,0),(0,0,1)$; also sign errors in the $2\times2$ minors when computing the …