let $p_{1}$ , be the plane $3 x - y - 7 z = 11$ and $p_{2}$ be the plane passing through the points (2,-1,0) ,(2,0,-1) and (5,1,1) .If the foot of the perpendicular drawn from the point (7,4,-1) on the line of intersection of the planes $p_{1}$ and $p_{2}$ is $(α, β, γ),$ then $(2 α + 3 β + 5 γ)$ is

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answer is 28.

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Detailed Solution

$\begin{array}{l} P_{1} : 3 x - y - 7 z = 11 \\ P_{2} : |\begin{matrix} x - 2 & y + 1 & z \\ 0 & 1 & - 1 \\ 3 & 2 & 1 \end{matrix}| = 0 \\ P_{2} : x - y - z = 3 \end{array}$

Eq. of line of intersection of $p_{1} & p_{2}$ is

$\frac{x - 4}{3} = \frac{y - 1}{2} = z$

Let N be the foot of A (7, 4, -1) then

$\begin{array}{l} N = (3 d + 4, 2 d + 1, d) \\ \bar{A N} . (3 \hat{i} + 2 \hat{j} + \hat{k}) = 0 \\ d = 1 \\ N = (α, β, γ) = (7, 3, 1) \\ 2 α + 3 β + 5 γ = 14 + 9 + 5 = 28 \end{array}$