The equation of the plane through (3, 1, -3)
and (1, -2, 2) and parallel 'to the line with $d . r' s$
1, 1,-2 is

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$x - y + z + 1 = 0$

$x + y - z + 1 = 0$

$x - y - z - 1 = 0$

$x + y + z - 1 = 0$

answer is D.

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Detailed Solution

$A (3, 1, - 3), B (1, - 2, 2) \Rightarrow$ the $d . r' s$ of $A B$ are -2, -3, 5 and d.r s of given line are 1, 1 -2.

The normal to the plane is perpendicular to the lines with $d . r' s$ -2, -3, 5 and 1, 1, -2.

The plane is $|\begin{array}{ccc} x - 3 & y - 1 & z + 3 \\ - 2 & - 3 & 5 \\ 1 & 1 & - 2 \end{array}| = 0$

$\Rightarrow (x - 3) 1 - (y - 1) (- 1) + (z + 3) 1 = 0 or x + y + z - 1 = 0 .$

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