l, m, n are three unit vectors in a right-han | vedclass

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(B) Let $l, m, n$ be $\hat{i}, \hat{j}, \hat{k}$.Points are $A(p, 7, -6), B(2, 5, -4), C(1, 4, -3)$.Since $A, B, C$ are collinear, the vector $\vec{AB

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$\frac{-40}{19}$

Vedclass · Answer

(B) Let $l, m, n$ be $\hat{i}, \hat{j}, \hat{k}$.Points are $A(p, 7, -6), B(2, 5, -4), C(1, 4, -3)$.Since $A, B, C$ are collinear, the vector $\vec{AB}$ must be parallel to $\vec{BC}$.$\vec{AB} = (2-p)\hat{i} - 2\hat{j} + 2\hat{k}$.$\vec{BC} = (1-2)\hat{i} + (4-5)\hat{j} + (-3+4)\hat{k} = -\hat{i} - \hat{j} + \hat{k}$.Since $\vec{AB} = k\vec{BC}$, we have $\frac{2-p}{-1} = \frac{-2}{-1} = \frac{2}{1} = 2$.Thus, $2-p = -2 \implies p = 4$.The point is $(-p, p, p+1) = (-4, 4, 5)$.The line $L$ passes through $B(2, 5, -4)$ and has direction vector $\vec{v} = (-1, -1, 1)$.The plane contains the line $L$ and the point $P(-4, 4, 5)$.The normal to the plane $\vec{n}$ is $\vec{PB} 	imes \vec{v}$.$\vec{PB} = (2 - (-4))\hat{i} + (5 - 4)\hat{j} + (-4 - 5)\hat{k} = 6\hat{i} + \hat{j} - 9\hat{k}$.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 6 & 1 & -9 \ -1 & -1 & 1 \end{vmatrix} = \hat{i}(1-9) - \hat{j}(6-9) + \hat{k}(-6+1) = -8\hat{i} + 3\hat{j} - 5\hat{k}$.The equation of the plane is $-8(x-2) + 3(y-5) - 5(z+4) = 0$.$-8x + 16 + 3y - 15 - 5z - 20 = 0 \implies -8x + 3y - 5z = 19$.Dividing by $19$, we get $-\frac{8}{19}x + \frac{3}{19}y - \frac{5}{19}z = 1$.Here $a = -\frac{8}{19}, b = \frac{3}{19}, c = -\frac{5}{19}$.$p(a+b+c) = 4 	imes (\frac{-8+3-5}{19}) = 4 	imes (\frac{-10}{19}) = -\frac{40}{19}$.

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