Let $P$ denotes plane containing all points equidistant from $A (- 4, 2, 1), B (2, - 4, 3)$ . $Q$ be the plane $x - y + cz = 1 where c \in R$ then

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Plane P is perpendicular to plane Q iff c =1

If angle between the planes P and Q is 45° then product of all values of c is -2

If the line L with equation $\frac{x - 1}{1} = \frac{y + 2}{3} = \frac{z - 7}{- 1}$ intersect the plane P at $(x_{0}, y_{0}, z_{0})$ then $x_{0} + y_{0} + z_{0} = 12$

The plane P is parallel to plane Q iff $c = \frac{1}{3}$

answer is A, B, C.

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Detailed Solution

Required planed passing through mid (AB) and perpendicular to AB
$\begin{array}{l} ∴ equation of P is 3 (x + 1) - 3 (y + 1) + 1 (z - 2) = 0 \\ P \equiv 3 x - 3 y + z - 2 = 0 \\ Q \equiv x - y + cz - 1 = 0 \\ P / / Q \Rightarrow \frac{3}{1} = \frac{1}{c} \Rightarrow c = \frac{1}{3} \\ \cos 45^{0} = \frac{\vec{n_{1}} \cdot \vec{n_{2}}}{|\vec{n_{1}}| |\vec{n_{2}}|} = |\frac{6 + c}{\sqrt{19} \sqrt{2 + c^{2}}}| \\ \Rightarrow 17 c^{2} - 24 c - 34 = 0 \Rightarrow c_{1} c_{2} = - 2 \\ \to x_{0} = 3, y_{0} = 4, z_{0} = 5 \Rightarrow x_{0} + y_{0} + z_{0} = 12 \end{array}$