Let the plane ax + by + cz = d pass through ( | vedclass

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(D) The normal vector $\vec{n}$ of the required plane is perpendicular to the normal vectors of the given planes,$\vec{n_1} = 2\hat{i} + \hat{j} - 5\h

Vedclass · Accepted Answer

$22$

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(D) The normal vector $\vec{n}$ of the required plane is perpendicular to the normal vectors of the given planes,$\vec{n_1} = 2\hat{i} + \hat{j} - 5\hat{k}$ and $\vec{n_2} = 3\hat{i} + 5\hat{j} - 7\hat{k}$.Thus,$\vec{n} = \vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -5 \ 3 & 5 & -7 \end{vmatrix} = \hat{i}(-7 + 25) - \hat{j}(-14 + 15) + \hat{k}(10 - 3) = 18\hat{i} - \hat{j} + 7\hat{k}$.The equation of the plane is $18x - y + 7z = d$.Since the plane passes through $(2, 3, -5)$,we have $18(2) - (3) + 7(-5) = d$,which gives $36 - 3 - 35 = d$,so $d = -2$.The equation of the plane is $18x - y + 7z = -2$,or $-18x + y - 7z = 2$.Comparing this with $ax + by + cz = d$,we get $a = -18, b = 1, c = -7, d = 2$.Since $	ext{gcd}(|-18|, |1|, |-7|, 2) = 1$ and $d > 0$,these values are correct.Finally,$a + 7b + c + 20d = -18 + 7(1) + (-7) + 20(2) = -18 + 7 - 7 + 40 = 22$.

Let the plane $ax + by + cz = d$ pass through $(2, 3, -5)$ and be perpendicular to the planes $2x + y - 5z = 10$ and $3x + 5y - 7z = 12$. If $a, b, c, d$ are integers,$d > 0$,and $\text{gcd}(|a|, |b|, |c|, d) = 1$,then the value of $a + 7b + c + 20d$ is equal to

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