If the equation of the plane containing the line x + 2y + 3z - 4 = 0, 2x + y - z + 5 = 0, and perpendicular to the plane

Question

If the equation of the plane containing the line $x + 2y + 3z - 4 = 0$ , $2x + y - z + 5 = 0$ , and perpendicular to the plane $\vec{r} = (\hat{i} - \hat{j}) + \lambda (\hat{i} + \hat{j} + \hat{k}) + \mu (\hat{i} - 2\hat{j} + 3\hat{k})$ is $ax + by + cz = 4$ , then $a - b + c$ is equal to:

A
$22$
B
$24$
C
$20$
D
$18$

Accepted Answer

Given: The required plane contains the line of intersection of $x + 2y + 3z - 4 = 0$ and $2x + y - z + 5 = 0$ . It is also perpendicular to the plane $\vec{r} = (\hat{i} - \hat{j}) + \lambda (\hat{i} + \hat{j} + \hat{k}) + \mu (\hat{i} - 2\hat{j} + 3\hat{k})$ .

Find: For the plane $ax + by + cz = 4$ , find $a - b + c$ .

From the solution, the direction ratios are obtained by taking the cross-product of the given vectors:

\vec{r}_1 = \hat{i} - \hat{j}, \quad \vec{r}_2 = \hat{i} + \hat{j} + \hat{k}, \quad \vec{r}_3 = \hat{i} - 2\hat{j} + 3\hat{k}

And

\vec{r}_1 \times \vec{r}_2 = (-27\hat{i} - 30\hat{j} - 25\hat{k})

A point on the plane is given as

\left(0, \frac{11}{5}, \frac{14}{5}\right)

Substituting this into the general equation, we get

27x + 30y + 25z = 4

Therefore,

a = 27, \quad b = 30, \quad c = 25

So,

a - b + c = 27 - 30 + 25 = 22

Therefore, the value of $a - b + c$ is $22$ . The correct option is A.

Given: the solution provides a point on the required plane as $\left(0, \frac{11}{5}, \frac{14}{5}\right)$ and the plane equation as $27x + 30y + 25z = 4$ .

Find: The value of $a - b + c$ .

Comparing

ax + by + cz = 4

with

27x + 30y + 25z = 4

we directly identify

a = 27, \quad b = 30, \quad c = 25

Now compute

a - b + c = 27 - 30 + 25 = 22

Hence, the required value is $22$ .

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