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$\mathbf{l}, \mathbf{m}, \mathbf{n}$ are three unit vectors in a right handed system and $L$ is a line through the points $A, B, C$ whose position vectors are $p \mathbf{l}+7 \mathbf{m}-6 \mathbf{n}, 2 \mathbf{l}+5 \mathbf{m}-4 \mathbf{n}$ and $\mathbf{l}+4 \mathbf{m}-3 \mathbf{n}$ respectively. If the equation of the plane containing $L$ and the points $(-p, p, p+1)$ is $a x+b y+c z=1$, then $p(a+b+c)=$
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Verified Answer
The correct answer is:
$\frac{-40}{19}$
Let $\mathbf{l}, \mathbf{m}, \mathbf{n}$ be $\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}$
$A(p \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}), B(2 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})$ and $C(\hat{\mathbf{i}}+4 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})$
Now, $A, B, C$ lie on line $L$
$\therefore \quad \mathbf{A B}=(2-p) \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$
$\mathbf{B C}=-\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$
$A, B, C$ are collinear
$\therefore \quad \frac{2-p}{-1}=\frac{-2}{-1}=\frac{2}{1}$
$2-p=-2 \Rightarrow p=4$
d.r. of line $(-1,-1,1)$
Equation of plane through part $(-p, p, p+1)$
i.e., $(-4,4,5)$
$\left|\begin{array}{ccc}x-2 & y-5 & z+4 \\ -4-2 & 4-5 & 5+4 \\ -1 & -1 & 1\end{array}\right|=0$
$\left|\begin{array}{ccc}x-2 & y-5 & z+4 \\ -6 & -1 & 9 \\ -1 & -1 & 1\end{array}\right|=0$
$(x-2)(-1+9)-(y-5)(-6+9)+(z+4)(6-1)=0$
$8 x-3 y+5 z=-19$
$-\frac{8}{19} x+\frac{3}{19} y-\frac{5}{19} z=1$
$\therefore \quad a=-\frac{8}{19}, b=\frac{3}{19}, c=\frac{-5}{19} \quad \because p(a+b+c)$
$a+b+c=-\frac{8}{19}+\frac{3}{19}-\frac{5}{19}=-\frac{10}{19} 4 \times \frac{-10}{19}=-\frac{40}{19}$
$A(p \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}), B(2 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})$ and $C(\hat{\mathbf{i}}+4 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})$
Now, $A, B, C$ lie on line $L$
$\therefore \quad \mathbf{A B}=(2-p) \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$
$\mathbf{B C}=-\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$
$A, B, C$ are collinear
$\therefore \quad \frac{2-p}{-1}=\frac{-2}{-1}=\frac{2}{1}$
$2-p=-2 \Rightarrow p=4$
d.r. of line $(-1,-1,1)$
Equation of plane through part $(-p, p, p+1)$
i.e., $(-4,4,5)$
$\left|\begin{array}{ccc}x-2 & y-5 & z+4 \\ -4-2 & 4-5 & 5+4 \\ -1 & -1 & 1\end{array}\right|=0$
$\left|\begin{array}{ccc}x-2 & y-5 & z+4 \\ -6 & -1 & 9 \\ -1 & -1 & 1\end{array}\right|=0$
$(x-2)(-1+9)-(y-5)(-6+9)+(z+4)(6-1)=0$
$8 x-3 y+5 z=-19$
$-\frac{8}{19} x+\frac{3}{19} y-\frac{5}{19} z=1$
$\therefore \quad a=-\frac{8}{19}, b=\frac{3}{19}, c=\frac{-5}{19} \quad \because p(a+b+c)$
$a+b+c=-\frac{8}{19}+\frac{3}{19}-\frac{5}{19}=-\frac{10}{19} 4 \times \frac{-10}{19}=-\frac{40}{19}$
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